给一根绳子 s ，任务是删除给定字符串中的所有重复项。 下面是删除字符串中重复项的不同方法。
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方法1（简单）
C++
// CPP program to remove duplicate character // from character array and print in sorted // order #include <bits/stdc++.h> using namespace std; char *removeDuplicate( char str[], int n) { // Used as index in the modified string int index = 0; // Traverse through all characters for ( int i=0; i<n; i++) { // Check if str[i] is present before it int j; for (j=0; j<i; j++) if (str[i] == str[j]) break ; // If not present, then add it to // result. if (j == i) str[index++] = str[i]; } return str; } // Driver code int main() { char str[]= "geeksforgeeks" ; int n = sizeof (str) / sizeof (str[0]); cout << removeDuplicate(str, n); return 0; } |
JAVA
// Java program to remove duplicate character // from character array and print in sorted // order import java.util.*; class GFG { static String removeDuplicate( char str[], int n) { // Used as index in the modified string int index = 0 ; // Traverse through all characters for ( int i = 0 ; i < n; i++) { // Check if str[i] is present before it int j; for (j = 0 ; j < i; j++) { if (str[i] == str[j]) { break ; } } // If not present, then add it to // result. if (j == i) { str[index++] = str[i]; } } return String.valueOf(Arrays.copyOf(str, index)); } // Driver code public static void main(String[] args) { char str[] = "geeksforgeeks" .toCharArray(); int n = str.length; System.out.println(removeDuplicate(str, n)); } } // This code is contributed by Rajput-Ji |
蟒蛇3
string = "geeksforgeeks" p = "" for char in string: if char not in p: p = p + char print (p) k = list ( "geeksforgeeks" ) |
C#
// C# program to remove duplicate character // from character array and print in sorted // order using System; using System.Collections.Generic; class GFG { static String removeDuplicate( char []str, int n) { // Used as index in the modified string int index = 0; // Traverse through all characters for ( int i = 0; i < n; i++) { // Check if str[i] is present before it int j; for (j = 0; j < i; j++) { if (str[i] == str[j]) { break ; } } // If not present, then add it to // result. if (j == i) { str[index++] = str[i]; } } char [] ans = new char [index]; Array.Copy(str, ans, index); return String.Join( "" , ans); } // Driver code public static void Main(String[] args) { char []str = "geeksforgeeks" .ToCharArray(); int n = str.Length; Console.WriteLine(removeDuplicate(str, n)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript program to remove duplicate character // from character array and print in sorted // order function removeDuplicate(str, n) { // Used as index in the modified string var index = 0; // Traverse through all characters for ( var i = 0; i < n; i++) { // Check if str[i] is present before it var j; for (j = 0; j < i; j++) { if (str[i] == str[j]) { break ; } } // If not present, then add it to // result. if (j == i) { str[index++] = str[i]; } } return str.join( "" ).slice(str, index); } // Driver code var str = "geeksforgeeks" .split( "" ); var n = str.length; document.write(removeDuplicate(str, n)); // This code is contributed by shivanisinghss2110 </script> |
输出：
geksfor
时间复杂性： O（n*n） 辅助空间： O（1） 保持元素的顺序与输入相同。
方法2（使用BST） 使用 设置 它实现了一个自平衡二叉搜索树。
C++
// CPP program to remove duplicate character // from character array and print in sorted // order #include <bits/stdc++.h> using namespace std; char *removeDuplicate( char str[], int n) { // create a set using string characters // excluding ' ' set< char >s (str, str+n-1); // print content of the set int i = 0; for ( auto x : s) str[i++] = x; str[i] = ' ' ; return str; } // Driver code int main() { char str[]= "geeksforgeeks" ; int n = sizeof (str) / sizeof (str[0]); cout << removeDuplicate(str, n); return 0; } |
JAVA
// Java program to remove duplicate character // from character array and print in sorted // order import java.util.*; class GFG { static void removeDuplicate( char str[], int n) { // Create a set using String characters // excluding ' ' HashSet<Character> s = new LinkedHashSet<>(n - 1 ); // HashSet doesn't allow repetition of elements for ( char x : str) s.add(x); // Print content of the set for ( char x : s) System.out.print(x); } // Driver code public static void main(String[] args) { char str[] = "geeksforgeeks" .toCharArray(); int n = str.length; removeDuplicate(str, n); } } // This code is contributed by todaysgaurav |
蟒蛇3
# Python program to remove duplicate character # from character array and print in sorted # order def removeDuplicate( str , n): s = set () # Create a set using String characters for i in str : s.add(i) # Print content of the set st = "" for i in s: st = st + i return st # Driver code str = "geeksforgeeks" n = len ( str ) print (removeDuplicate( list ( str ), n)) # This code is contributed by rajsanghavi9. |
C#
// C# program to remove duplicate character // from character array and print in sorted // order using System; using System.Collections.Generic; public class GFG{ static char []removeDuplicate( char []str, int n) { // Create a set using String characters // excluding ' ' HashSet< char >s = new HashSet< char >(n - 1); foreach ( char x in str) s.Add(x); char [] st = new char [s.Count]; // Print content of the set int i = 0; foreach ( char x in s) st[i++] = x; return st; } // Driver code public static void Main(String[] args) { char []str= "geeksforgeeks" .ToCharArray(); int n = str.Length; Console.Write(removeDuplicate(str, n)); } } // This code contributed by gauravrajput1 |
Javascript
<script> // javascript program to remove duplicate character // from character array and print in sorted // order function removeDuplicate( str , n) { // Create a set using String characters // excluding ' ' var s = new Set(); // HashSet doesn't allow repetition of elements for ( var i = 0;i<n;i++) s.add(str[i]); // Print content of the set for (const v of s) { document.write(v); } } // Driver code var str = "geeksforgeeks" ; var n = str.length; removeDuplicate(str, n); // This code is contributed by umadevi9616 </script> |
输出：
efgkors
时间复杂性 ：O（n日志n） 辅助空间 ：O（n）
幸亏 阿尼维斯·蒂瓦里 感谢你提出这种方法。
它不保持元素的顺序与输入相同，而是按排序顺序打印它们。
方法3（使用排序） 算法：
1) Sort the elements. 2) Now in a loop, remove duplicates by comparing the current character with previous character. 3) Remove extra characters at the end of the resultant string.
例子：
Input string: geeksforgeeks1) Sort the characters eeeefggkkorss2) Remove duplicates efgkorskkorss3) Remove extra characters efgkors
请注意，此方法不会保持输入字符串的原始顺序。例如，如果我们要删除geeksforgek的重复项并保持字符顺序不变，那么输出应该是geksfor，但上面的函数返回efgkos。我们可以通过存储原始订单来修改此方法。
实施：
C++
// C++ program to remove duplicates, the order of // characters is not maintained in this program #include<bits/stdc++.h> using namespace std; /* Function to remove duplicates in a sorted array */ char *removeDupsSorted( char *str) { int res_ind = 1, ip_ind = 1; /* In place removal of duplicate characters*/ while (*(str + ip_ind)) { if (*(str + ip_ind) != *(str + ip_ind - 1)) { *(str + res_ind) = *(str + ip_ind); res_ind++; } ip_ind++; } /* After above step string is efgkorskkorss. Removing extra kkorss after string*/ *(str + res_ind) = ' ' ; return str; } /* Function removes duplicate characters from the string This function work in-place and fills null characters in the extra space left */ char *removeDups( char *str) { int n = strlen (str); // Sort the character array sort(str, str+n); // Remove duplicates from sorted return removeDupsSorted(str); } /* Driver program to test removeDups */ int main() { char str[] = "geeksforgeeks" ; cout << removeDups(str); return 0; } |
C
// C++ program to remove duplicates, the order of // characters is not maintained in this program # include <stdio.h> # include <stdlib.h> # include <string.h> /* Function to remove duplicates in a sorted array */ char *removeDupsSorted( char *str); /* Utility function to sort array A[] */ void quickSort( char A[], int si, int ei); /* Function removes duplicate characters from the string This function work in-place and fills null characters in the extra space left */ char *removeDups( char *str) { int len = strlen (str); quickSort(str, 0, len-1); return removeDupsSorted(str); } /* Function to remove duplicates in a sorted array */ char *removeDupsSorted( char *str) { int res_ind = 1, ip_ind = 1; /* In place removal of duplicate characters*/ while (*(str + ip_ind)) { if (*(str + ip_ind) != *(str + ip_ind - 1)) { *(str + res_ind) = *(str + ip_ind); res_ind++; } ip_ind++; } /* After above step string is efgkorskkorss. Removing extra kkorss after string*/ *(str + res_ind) = ' ' ; return str; } /* Driver program to test removeDups */ int main() { char str[] = "geeksforgeeks" ; printf ( "%s" , removeDups(str)); getchar (); return 0; } /* FOLLOWING FUNCTIONS ARE ONLY FOR SORTING PURPOSE */ void exchange( char *a, char *b) { char temp; temp = *a; *a = *b; *b = temp; } int partition( char A[], int si, int ei) { char x = A[ei]; int i = (si - 1); int j; for (j = si; j <= ei - 1; j++) { if (A[j] <= x) { i++; exchange(&A[i], &A[j]); } } exchange (&A[i + 1], &A[ei]); return (i + 1); } /* Implementation of Quick Sort A[] --> Array to be sorted si --> Starting index ei --> Ending index */ void quickSort( char A[], int si, int ei) { int pi; /* Partitioning index */ if (si < ei) { pi = partition(A, si, ei); quickSort(A, si, pi - 1); quickSort(A, pi + 1, ei); } } |
JAVA
// Java program to remove duplicates, the order of // characters is not maintained in this program import java.util.Arrays; public class GFG { /* Method to remove duplicates in a sorted array */ static String removeDupsSorted(String str) { int res_ind = 1 , ip_ind = 1 ; // Character array for removal of duplicate characters char arr[] = str.toCharArray(); /* In place removal of duplicate characters*/ while (ip_ind != arr.length) { if (arr[ip_ind] != arr[ip_ind- 1 ]) { arr[res_ind] = arr[ip_ind]; res_ind++; } ip_ind++; } str = new String(arr); return str.substring( 0 ,res_ind); } /* Method removes duplicate characters from the string This function work in-place and fills null characters in the extra space left */ static String removeDups(String str) { // Sort the character array char temp[] = str.toCharArray(); Arrays.sort(temp); str = new String(temp); // Remove duplicates from sorted return removeDupsSorted(str); } // Driver Method public static void main(String[] args) { String str = "geeksforgeeks" ; System.out.println(removeDups(str)); } } |
蟒蛇3
# Python program to remove duplicates, the order of # characters is not maintained in this program # Utility function to convert string to list def toMutable(string): temp = [] for x in string: temp.append(x) return temp # Utility function to convert string to list def toString( List ): return ''.join( List ) # Function to remove duplicates in a sorted array def removeDupsSorted( List ): res_ind = 1 ip_ind = 1 # In place removal of duplicate characters while ip_ind ! = len ( List ): if List [ip_ind] ! = List [ip_ind - 1 ]: List [res_ind] = List [ip_ind] res_ind + = 1 ip_ind + = 1 # After above step string is efgkorskkorss. # Removing extra kkorss after string string = toString( List [ 0 :res_ind]) return string # Function removes duplicate characters from the string # This function work in-place and fills null characters # in the extra space left def removeDups(string): # Convert string to list List = toMutable(string) # Sort the character list List .sort() # Remove duplicates from sorted return removeDupsSorted( List ) # Driver program to test the above functions string = "geeksforgeeks" print (removeDups(string)) # This code is contributed by Bhavya Jain |
C#
// C# program to remove duplicates, the order of // characters is not maintained in this program using System; class GFG { /* Method to remove duplicates in a sorted array */ static String removeDupsSorted(String str) { int res_ind = 1, ip_ind = 1; // Character array for removal of duplicate characters char []arr = str.ToCharArray(); /* In place removal of duplicate characters*/ while (ip_ind != arr.Length) { if (arr[ip_ind] != arr[ip_ind-1]) { arr[res_ind] = arr[ip_ind]; res_ind++; } ip_ind++; } str = new String(arr); return str.Substring(0,res_ind); } /* Method removes duplicate characters from the string This function work in-place and fills null characters in the extra space left */ static String removeDups(String str) { // Sort the character array char []temp = str.ToCharArray(); Array.Sort(temp); str = String.Join( "" ,temp); // Remove duplicates from sorted return removeDupsSorted(str); } // Driver Method public static void Main(String[] args) { String str = "geeksforgeeks" ; Console.WriteLine(removeDups(str)); } } // This code is contributed by 29AjayKumar |
Javascript
<script> function removeDuplicate(string) { return string.split( '' ) .filter( function (item, pos, self) { return self.indexOf(item) == pos; } ).join( '' ); } var str = "geeksforgeeks" ; document.write( " " +removeDuplicate(str)); //This code is contributed by SoumikMondal </script> |
输出：
efgkors
时间复杂性： O（n logn）如果我们使用一些nlogn排序算法而不是快速排序。
辅助空间： O（1）
方法4（使用哈希）
算法：
1: Initialize: str = "test string" /* input string */ ip_ind = 0 /* index to keep track of location of next character in input string */ res_ind = 0 /* index to keep track of location of next character in the resultant string */ bin_hash[0..255] = {0,0, ….} /* Binary hash to see if character is already processed or not */2: Do following for each character *(str + ip_ind) in input string: (a) if bin_hash is not set for *(str + ip_ind) then // if program sees the character *(str + ip_ind) first time (i) Set bin_hash for *(str + ip_ind) (ii) Move *(str + ip_ind) to the resultant string. This is done in-place. (iii) res_ind++ (b) ip_ind++ /* String obtained after this step is "the stringing" */3: Remove extra characters at the end of the resultant string. /* String obtained after this step is "te string" */
实施：
C++
#include <bits/stdc++.h> using namespace std; # define NO_OF_CHARS 256 # define bool int /* Function removes duplicate characters from the string This function work in-place and fills null characters in the extra space left */ char *removeDups( char str[]) { bool bin_hash[NO_OF_CHARS] = {0}; int ip_ind = 0, res_ind = 0; char temp; /* In place removal of duplicate characters*/ while (*(str + ip_ind)) { temp = *(str + ip_ind); if (bin_hash[temp] == 0) { bin_hash[temp] = 1; *(str + res_ind) = *(str + ip_ind); res_ind++; } ip_ind++; } /* After above step string is stringiittg. Removing extra iittg after string*/ *(str+res_ind) = ' ' ; return str; } /* Driver code */ int main() { char str[] = "geeksforgeeks" ; cout << removeDups(str); return 0; } // This code is contributed by rathbhupendra |
C
# include <stdio.h> # include <stdlib.h> # define NO_OF_CHARS 256 # define bool int /* Function removes duplicate characters from the string This function work in-place and fills null characters in the extra space left */ char *removeDups( char *str) { bool bin_hash[NO_OF_CHARS] = {0}; int ip_ind = 0, res_ind = 0; char temp; /* In place removal of duplicate characters*/ while (*(str + ip_ind)) { temp = *(str + ip_ind); if (bin_hash[temp] == 0) { bin_hash[temp] = 1; *(str + res_ind) = *(str + ip_ind); res_ind++; } ip_ind++; } /* After above step string is stringiittg. Removing extra iittg after string*/ *(str+res_ind) = ' ' ; return str; } /* Driver program to test removeDups */ int main() { char str[] = "geeksforgeeks" ; printf ( "%s" , removeDups(str)); getchar (); return 0; } |
JAVA
// Java program to remove duplicates import java.util.*; class RemoveDuplicates { /* Function removes duplicate characters from the string This function work in-place */ void removeDuplicates(String str) { LinkedHashSet<Character> lhs = new LinkedHashSet<>(); for ( int i= 0 ;i<str.length();i++) lhs.add(str.charAt(i)); // print string after deleting duplicate elements for (Character ch : lhs) System.out.print(ch); } /* Driver program to test removeDuplicates */ public static void main(String args[]) { String str = "geeksforgeeks" ; RemoveDuplicates r = new RemoveDuplicates(); r.removeDuplicates(str); } } // This code has been contributed by Amit Khandelwal (Amit Khandelwal 1) |
蟒蛇3
# Python program to remove duplicate characters from an # input string NO_OF_CHARS = 256 # Since strings in Python are immutable and cannot be changed # This utility function will convert the string to list def toMutable(string): List = [] for i in string: List .append(i) return List # Utility function that changes list to string def toString( List ): return ''.join( List ) # Function removes duplicate characters from the string # This function work in-place and fills null characters # in the extra space left def removeDups(string): bin_hash = [ 0 ] * NO_OF_CHARS ip_ind = 0 res_ind = 0 temp = '' mutableString = toMutable(string) # In place removal of duplicate characters while ip_ind ! = len (mutableString): temp = mutableString[ip_ind] if bin_hash[ ord (temp)] = = 0 : bin_hash[ ord (temp)] = 1 mutableString[res_ind] = mutableString[ip_ind] res_ind + = 1 ip_ind + = 1 # After above step string is stringiittg. # Removing extra iittg after string return toString(mutableString[ 0 :res_ind]) # Driver program to test the above functions string = "geeksforgeeks" print (removeDups(string)) # A shorter version for this program is as follows # import collections # print ''.join(collections.OrderedDict.fromkeys(string)) # This code is contributed by Bhavya Jain |
C#
// C# program to remove duplicates using System; using System.Collections.Generic; class GFG { /* Function removes duplicate characters from the string. This function work in-place */ void removeDuplicates(String str) { HashSet< char > lhs = new HashSet< char >(); for ( int i = 0; i < str.Length; i++) lhs.Add(str[i]); // print string after deleting // duplicate elements foreach ( char ch in lhs) Console.Write(ch); } // Driver Code public static void Main(String []args) { String str = "geeksforgeeks" ; GFG r = new GFG(); r.removeDuplicates(str); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // javascript program to remove duplicates /* * Function removes duplicate characters from the string This function work * in-place */ function removeDuplicates( str) { var lhs = new Set(); for ( var i = 0; i < str.length; i++) lhs.add(str[i]); // print string after deleting duplicate elements for ( var ch of lhs) document.write(ch); } /* Driver program to test removeDuplicates */ var str = "geeksforgeeks" ; removeDuplicates(str); // This code is contributed by umadevi9616 </script> |
输出：
geksfor
时间复杂性： O（n）
要点：
- 方法2不将字符保留为原始字符串，但方法4保留。
- 假设输入字符串中可能包含的字符数为256。不应相应地更改任何字符。
- 计数数组（count）的内存分配使用calloc（）而不是malloc（），以将分配的内存初始化为“”。也可以使用malloc（）后跟memset（）。
- 如果给定了数组中整数的范围，上述算法也适用于整数数组输入。一个示例问题是，如果输入数组只包含1000到1100之间的整数，则查找输入数组中出现的最大数
方法5 （使用 IndexOf（） 方法）： 先决条件： JAVA IndexOf（） 方法
C++
// C++ program to create a unique string #include <bits/stdc++.h> using namespace std; // Function to make the string unique string unique(string s) { string str; int len = s.length(); // loop to traverse the string and // check for repeating chars using // IndexOf() method in Java for ( int i = 0; i < len; i++) { // character at i'th index of s char c = s[i]; // If c is present in str, it returns // the index of c, else it returns npos auto found = str.find(c); if (found == std::string::npos) { // Adding c to str if npos is returned str += c; } } return str; } // Driver code int main() { // Input string with repeating chars string s = "geeksforgeeks" ; cout << unique(s) << endl; } // This code is contributed by nirajgusain5 |
JAVA
// Java program to create a unique string import java.util.*; class IndexOf { // Function to make the string unique public static String unique(String s) { String str = new String(); int len = s.length(); // loop to traverse the string and // check for repeating chars using // IndexOf() method in Java for ( int i = 0 ; i < len; i++) { // character at i'th index of s char c = s.charAt(i); // if c is present in str, it returns // the index of c, else it returns -1 if (str.indexOf(c) < 0 ) { // adding c to str if -1 is returned str += c; } } return str; } // Driver code public static void main(String[] args) { // Input string with repeating chars String s = "geeksforgeeks" ; System.out.println(unique(s)); } } |
蟒蛇3
# Python 3 program to create a unique string # Function to make the string unique def unique(s): st = "" length = len (s) # loop to traverse the string and # check for repeating chars using # IndexOf() method in Java for i in range (length): # character at i'th index of s c = s[i] # if c is present in str, it returns # the index of c, else it returns - 1 # print(st.index(c)) if c not in st: # adding c to str if -1 is returned st + = c return st # Driver code if __name__ = = "__main__" : # Input string with repeating chars s = "geeksforgeeks" print (unique(s)) # This code is contributed by ukasp. |
C#
// C# program to create a unique string using System; public class IndexOf { // Function to make the string unique public static String unique(String s) { String str = "" ; int len = s.Length; // loop to traverse the string and // check for repeating chars using // IndexOf() method in Java for ( int i = 0; i < len; i++) { // character at i'th index of s char c = s[i]; // if c is present in str, it returns // the index of c, else it returns -1 if (str.IndexOf(c) < 0) { // adding c to str if -1 is returned str += c; } } return str; } // Driver code public static void Main(String[] args) { // Input string with repeating chars String s = "geeksforgeeks" ; Console.WriteLine(unique(s)); } } // This code is contributed by Princi Singh |
Javascript
<script> // JavaScript program to create a unique string // Function to make the string unique function unique(s) { let str = "" ; let len = s.length; // loop to traverse the string and // check for repeating chars using // IndexOf() method in Java for (let i = 0; i < len; i++) { // character at i'th index of s let c = s[i]; // if c is present in str, it returns // the index of c, else it returns -1 if (str.indexOf(c) < 0) { // adding c to str if -1 is returned str += c; } } return str; } // Input string with repeating chars let s = "geeksforgeeks" ; document.write(unique(s)); </script> |
输出：
geksfor
谢谢 debjitdbb 感谢你提出这种方法。
方法6 （使用 无序地图STL 方法）： 先决条件： 无序C++映射STL C++方法
C++
// C++ program to create a unique string using unordered_map /* access time in unordered_map on is O(1) generally if no collisions occur and therefore it helps us check if an element exists in a string in O(1) time complexity with constant space. */ #include <bits/stdc++.h> using namespace std; char * removeDuplicates( char *s, int n){ unordered_map< char , int > exists; int index = 0; for ( int i=0;i<n;i++){ if (exists[s[i]]==0) { s[index++] = s[i]; exists[s[i]]++; } } return s; } //driver code int main(){ char s[] = "geeksforgeeks" ; int n = sizeof (s)/ sizeof (s[0]); cout<<removeDuplicates(s,n)<<endl; return 0; } |
JAVA
// Java program to create a unique String using unordered_map /* access time in unordered_map on is O(1) generally if no collisions occur and therefore it helps us check if an element exists in a String in O(1) time complexity with constant space. */ import java.util.*; class GFG{ static char [] removeDuplicates( char []s, int n){ Map<Character,Integer> exists = new HashMap<>(); String st = "" ; for ( int i = 0 ; i < n; i++){ if (!exists.containsKey(s[i])) { st += s[i]; exists.put(s[i], 1 ); } } return st.toCharArray(); } // driver code public static void main(String[] args){ char s[] = "geeksforgeeks" .toCharArray(); int n = s.length; System.out.print(removeDuplicates(s,n)); } } // This code is contributed by gauravrajput1 |
蟒蛇3
# Python program to create a unique string using unordered_map # access time in unordered_map on is O(1) generally if no collisions occur # and therefore it helps us check if an element exists in a string in O(1) # time complexity with constant space. def removeDuplicates(s, n): exists = {} index = 0 ans = "" for i in range ( 0 , n): if s[i] not in exists or exists[s[i]] = = 0 : s[index] = s[i] print (s[index], end = '') index + = 1 exists[s[i]] = 1 # driver code s = "geeksforgeeks" s1 = list (s) n = len (s1) removeDuplicates(s1, n) # This code is contributed by Palak Gupta |
C#
// C# program to create a unique String using unordered_map /* access time in unordered_map on is O(1) generally if no collisions occur and therefore it helps us check if an element exists in a String in O(1) time complexity with constant space. */ using System; using System.Collections.Generic; public class GFG{ static char [] removeDuplicates( char []s, int n){ Dictionary< char , int > exists = new Dictionary< char , int >(); String st = "" ; for ( int i = 0; i < n; i++){ if (!exists.ContainsKey(s[i])) { st += s[i]; exists.Add(s[i], 1); } } return st.ToCharArray(); } // driver code public static void Main(String[] args){ char []s = "geeksforgeeks" .ToCharArray(); int n = s.Length; Console.Write(removeDuplicates(s,n)); } } // This code is contributed by umadevi9616 |
Javascript
<script> // javascript program to create a unique String using unordered_map /* access time in unordered_map on is O(1) generally if no collisions occur and therefore it helps us check if an element exists in a String in O(1) time complexity with constant space. */ function removeDuplicates( s , n) { var exists = new Map(); var st = "" ; for ( var i = 0; i < n; i++) { if (!exists.has(s[i])) { st += s[i]; exists.set(s[i], 1); } } return st; } // driver code var s = "geeksforgeeks" ; var n = s.length; document.write(removeDuplicates(s, n)); // This code contributed by umadevi9616 </script> |
输出：
geksfor
时间复杂性： O（n） 辅助空间： O（n） 谢谢 艾伦·詹姆斯·维诺伊 感谢你提出这种方法。
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