给定一个数组arr[]，数组的大小为n，另一个键为x，并给出一个段大小k。任务是在arr[]中的每个大小为k的段中找到键x。 例如：
null
输入： arr[]={3,5,2,4,9,3,1,7,3,11,12,3} x=3 k=3 输出： 对 数组中有4个大小为k的不重叠段，{3,5,2}，{4,9,3}，{1,7,3}和{11,12,3}。3.所有环节都有。 输入： arr[]={21,23,56,65,34,54,76,32,23,45,21,23,25} x=23 k=5 输出： 对 有三段，最后一段不是完整的{21,23,56,65,34}，{54,76,32,23,45}和{21,23,25}。 23号在所有窗口。 输入： arr[]={5,8,7,12,14,3,9} x=8 k=2 输出： 不
这个想法很简单，我们考虑每一个大小的k段，检查是否存在于窗口中。我们需要小心处理最后一部分。 以下是上述方法的实施情况：
C++
// C++ code to find the every segment size of // array have a search key x #include <bits/stdc++.h> using namespace std; bool findxinkwindowSize( int arr[], int x, int k, int n) { int i; for (i = 0; i < n; i = i + k) { // Search x in segment starting // from index i. int j; for (j = 0; j < k; j++) if (arr[i + j] == x) break ; // If loop didn't break if (j == k) return false ; } // If n is a multiple of k if (i == n) return true ; // Check in last segment if n // is not multiple of k. int j; for (j=i-k; j<n; j++) if (arr[j] == x) break ; if (j == n) return false ; return true ; } // main driver int main() { int arr[] = { 3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3 }; int x = 3, k = 3; int n = sizeof (arr) / sizeof (arr[0]); if (findxinkwindowSize(arr, x, k, n)) cout << "Yes" << endl; else cout << "No" << endl; return 0; } |
JAVA
// Java code to find the every // segment size of array have // a search key x import java.util.*; class GFG { static boolean findxinkwindowSize( int N, int [] arr, int x, int k) { int i; boolean b = false ; // Iterate from 0 to N - 1 for (i = 0 ; i < N; i = i + k) { // Iterate from 0 to k - 1 for ( int j = 0 ; j < k; j++) { if (i + j < N && arr[i + j] == x) break ; if (j == k) return false ; if (i + j >= N) return false ; } } if (i >= N) return true ; else return b; } // Driver Code public static void main(String args[]) { int arr[] = new int [] { 3 , 5 , 2 , 4 , 9 , 3 , 1 , 7 , 3 , 11 , 12 , 3 }; int x = 3 , k = 3 ; int n = arr.length; if (findxinkwindowSize(n, arr, x, k)) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by Vivek258709 |
Python 3
# Python 3 program to find # the every segment size of # array have a search key x def findxinkwindowSize(arr, x, k, n) : i = 0 while i < n : j = 0 # Search x in segment # starting from index i while j < k : if arr[i + j] = = x : break j + = 1 # If loop didn't break if j = = k : return False i + = k # If n is a multiple of k if i = = n : return True j = i - k # Check in last segment if n # is not multiple of k. while j < n : if arr[j] = = x : break j + = 1 if j = = n : return False return True # Driver Code if __name__ = = "__main__" : arr = [ 3 , 5 , 2 , 4 , 9 , 3 , 1 , 7 , 3 , 11 , 12 , 3 ] x, k = 3 , 3 n = len (arr) if (findxinkwindowSize(arr, x, k, n)) : print ( "Yes" ) else : print ( "No" ) # This code is contributed # by ANKITRAI1 |
C#
// C# code to find the every // segment size of array have // a search key x using System; class GFG { static bool findxinkwindowSize( int [] arr, int x, int k, int n) { int i; for (i = 0; i < n; i = i + k) { // Search x in segment // starting from index i. int j; for (j = 0; j < k; j++) if (arr[i + j] == x) break ; // If loop didn't break if (j == k) return false ; } // If n is a multiple of k if (i == n) return true ; // Check in last segment if // n is not multiple of k. int l; for (l = i - k; l < n; l++) if (arr[l] == x) break ; if (l == n) return false ; return true ; } // Driver Code public static void Main() { int [] arr = new int [] {3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3}; int x = 3, k = 3; int n = arr.Length; if (findxinkwindowSize(arr, x, k, n)) Console.Write( "Yes" ); else Console.Write( "No" ); } } // This code is contributed by ChitraNayal |
PHP
<?php // PHP code to find the every // segment size of array have // a search key x function findxinkwindowSize(& $arr , $x , $k , $n ) { for ( $i = 0; $i < $n ; $i = $i + $k ) { // Search x in segment // starting from index i. for ( $j = 0; $j < $k ; $j ++) if ( $arr [ $i + $j ] == $x ) break ; // If loop didn't break if ( $j == $k ) return false; } // If n is a multiple of k if ( $i == $n ) return true; // Check in last segment if n // is not multiple of k. for ( $j = $i - $k ; $j < $n ; $j ++) if ( $arr [ $j ] == $x ) break ; if ( $j == $n ) return false; return true; } // Driver Code $arr = array (3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3); $x = 3; $k = 3; $n = sizeof( $arr ); if (findxinkwindowSize( $arr , $x , $k , $n )) echo "Yes" ; else echo "No" ; // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script> // JavaScript code to find the every segment size of // array have a search key x function findxinkwindowSize( arr, x, k, n) { let i; for (i = 0; i < n; i = i + k) { // Search x in segment starting // from index i. let j; for (j = 0; j < k; j++) if (arr[i + j] == x) break ; // If loop didn't break if (j == k) return false ; } // If n is a multiple of k if (i == n) return true ; // Check in last segment if n // is not multiple of k. let j; for (j=i-k; j<n; j++) if (arr[j] == x) break ; if (j == n) return false ; return true ; } // main driver let arr = [ 3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3 ]; let x = 3, k = 3; let n = arr.length; if (findxinkwindowSize(arr, x, k, n)) document.write( "Yes" ); else document.write( "No" ); // This code contributed by aashish1995 </script> |
输出：
Yes
时间复杂度：O（n）
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