给定一个由N个元素组成的数组,通过删除除最后一个元素外的所有元素引用,以与给定相同的相对顺序打印元素。 例子 :
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输入: a[]={1,5,5,1,6,1} 输出: 5 6 1 删除位置1和4中的两个整数1。此外,删除位于位置2的整数5。 因此,左边的数组是{5,6,1} 输入: a[]={2,5,5,2} 输出: 5 2
方法:
- 散列每个元素的最后一次出现。
- 迭代N个元素的数组,如果元素的索引是散列的,那么打印数组元素。
以下是上述方法的实施情况:
C++
// C++ program to print the last occurrence // of every element in relative order #include <bits/stdc++.h> using namespace std; // Function to print the last occurrence // of every element in an array void printLastOccurrence( int a[], int n) { // used in hashing unordered_map< int , int > mp; // iterate and store the last index // of every element for ( int i = 0; i < n; i++) mp[a[i]] = i; // iterate and check for the last // occurrence of every element for ( int i = 0; i < n; i++) { if (mp[a[i]] == i) cout << a[i] << " " ; } } // Driver Code int main() { int a[] = { 1, 5, 5, 1, 6, 1 }; int n = sizeof (a) / sizeof (a[0]); printLastOccurrence(a, n); return 0; } |
JAVA
// Java program to print the // last occurrence of every // element in relative order import java.util.*; class GFG { // Function to print the last // occurrence of every element // in an array public static void printLastOccurrence( int a[], int n) { HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); // iterate and store the last // index of every element for ( int i = 0 ; i < n; i++) map.put(a[i], i); for ( int i = 0 ; i < n; i++) { if (map.get(a[i]) == i) System.out.print(a[i] + " " ); } } // Driver Code public static void main (String[] args) { int a[] = { 1 , 5 , 5 , 1 , 6 , 1 }; int n = a.length; printLastOccurrence(a, n); } } // This code is contributed // by ankita_saini |
Python3
# Python 3 program to print the last occurrence # of every element in relative order # Function to print the last occurrence # of every element in an array def printLastOccurrence(a, n): # used in hashing mp = {i: 0 for i in range ( 7 )} # iterate and store the last # index of every element for i in range (n): mp[a[i]] = i # iterate and check for the last # occurrence of every element for i in range (n): if (mp[a[i]] = = i): print (a[i], end = " " ) # Driver Code if __name__ = = '__main__' : a = [ 1 , 5 , 5 , 1 , 6 , 1 ] n = len (a) printLastOccurrence(a, n) # This code is contributed by # Surendra_Gangwar |
C#
// C# program to print the // last occurrence of every // element in relative order using System; class GFG { // Function to print the last // occurrence of every element // in an array public static void printLastOccurrence( int [] a, int n) { HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); // iterate and store the last // index of every element for ( int i = 0; i < n; i++) map.put(a[i], i); for ( int i = 0; i < n; i++) { if (map. get (a[i]) == i) Console.Write(a[i] + " " ); } } // Driver Code public static void Main () { int [] a = { 1, 5, 5, 1, 6, 1 }; int n = a.Length; printLastOccurrence(a, n); } } // This code is contributed // by ChitraNayal |
PHP
<?php // PHP program to print the last // occurrence of every element // in relative order // Function to print the last // occurrence of every element // in an array function printLastOccurrence(& $a , $n ) { // used in hashing $mp = array (); // iterate and store the last // index of every element for ( $i = 0; $i < $n ; $i ++) $mp [ $a [ $i ]] = $i ; // iterate and check for the last // occurrence of every element for ( $i = 0; $i < $n ; $i ++) { if ( $mp [ $a [ $i ]] == $i ) echo $a [ $i ] . " " ; } } // Driver Code $a = array (1, 5, 5, 1, 6, 1); $n = sizeof( $a ); printLastOccurrence( $a , $n ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to print the last // occurrence of every element // in relative order // Function to print the last // occurrence of every element // in an array function printLastOccurrence(a, n) { // used in hashing let mp = []; // iterate and store the last // index of every element for (let i = 0; i < n; i++) mp[a[i]] = i; // iterate and check for the last // occurrence of every element for (let i = 0; i < n; i++) { if (mp[a[i]] == i) document.write( a[i] + " " ); } } // Driver Code let a = [1, 5, 5, 1, 6, 1]; let n = a.length; printLastOccurrence(a, n); // This code is contributed by sravan kumar </script> |
输出:
5 6 1
时间复杂性: O(N*logn)
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