检查(I，j)是否存在，使得 arr[i]！= arr[j]，arr[arr[i]]等于 arr[arr[j]]
原文:https://www . geeksforgeeks . org/check-if-ij-exists-so-arri-arrj-and-arr RRI-等于-arr rrj/
给定数组 A[]。任务是确定是否有可能选择两个指数‘I’和‘j’，从而满足以下条件:-
- A [I] is not equal to A [J] .
- A[A[i]] 等于 A[A[j]] 。
注意:数组中元素的值小于 N 的值，即每 I，arr[i] < N. 示例:
Input: N = 4, A[] = {1, 1, 2, 3}
Output: Yes
As A[3] != to A[1] but A[A[3]] == A[A[1]]
Input: N = 4, A[] = {2, 1, 3, 3}
Output: No
As A[A[3]] == A[A[4]] but A[3] == A[4]
进场:
- 通过运行两个循环开始遍历 arr[]数组。
- 变量 i 指向指数 0，变量 j 指向下一个 I
- 如果 Arr[i]不等于 Arr[j]，则检查 Arr[Arr[I]–1]是否等于 Arr[Arr[j]–1]。如果是，则返回真。 否则也检查其他索引的 Arr[Arr[i]- 1]和 Arr[Arr[j]–1]。
- 重复以上步骤，直到遍历完所有元素/索引。
- 如果没有发现这样的指数，返回假。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that will tell whether
// such Indices present or Not.
bool checkIndices(int Arr[], int N)
{
for (int i = 0; i < N - 1; i++) {
for (int j = i + 1; j < N; j++) {
// Checking 1st condition i.e whether
// Arr[i] equal to Arr[j] or not
if (Arr[i] != Arr[j]) {
// Checking 2nd condition i.e whether
// Arr[Arr[i]] equal to Arr[Arr[j]] or not.
if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])
return true;
}
}
}
return false;
}
// Driver Code
int main()
{
int Arr[] = { 3, 2, 1, 1, 4 };
int N = sizeof(Arr) / sizeof(Arr[0]);
// Calling function.
checkIndices(Arr, N) ? cout << "Yes"
: cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
// Function that calculates marks.
class GFG
{
static boolean checkIndices(int Arr[], int N)
{
for (int i = 0; i < N - 1; i++) {
for (int j = i + 1; j < N; j++) {
// Checking 1st condition i.e whether
// Arr[i] equal to Arr[j] or not
if (Arr[i] != Arr[j]) {
// Checking 2nd condition i.e whether
// Arr[Arr[i]] equal to Arr[Arr[j]] or not.
if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])
return true;
}
}
}
return false;
}
// Driver code
public static void main(String args[])
{
int Arr[] = { 3, 2, 1, 1, 4 };
int N = Arr.length;
if(checkIndices(Arr, N))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is Contributed by
// Naman_Garg
Python 3
# Python 3 implementation of the
# above approach
# Function that will tell whether
# such Indices present or Not.
def checkIndices(Arr, N):
for i in range(N - 1):
for j in range(i + 1, N):
# Checking 1st condition i.e whether
# Arr[i] equal to Arr[j] or not
if (Arr[i] != Arr[j]):
# Checking 2nd condition i.e whether
# Arr[Arr[i]] equal to Arr[Arr[j]] or not.
if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1]):
return True
return False
# Driver Code
if __name__ == "__main__":
Arr = [ 3, 2, 1, 1, 4 ]
N =len(Arr)
# Calling function.
if checkIndices(Arr, N):
print("Yes")
else:
print("No")
# This code is contributed by ita_c
C
// C# implementation of the above approach
using System;
class GFG
{
// Function that calculates marks.
static bool checkIndices(int []Arr, int N)
{
for (int i = 0; i < N - 1; i++)
{
for (int j = i + 1; j < N; j++)
{
// Checking 1st condition i.e whether
// Arr[i] equal to Arr[j] or not
if (Arr[i] != Arr[j])
{
// Checking 2nd condition i.e
// whether Arr[Arr[i]] equal
// to Arr[Arr[j]] or not.
if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])
return true;
}
}
}
return false;
}
// Driver code
static public void Main ()
{
int []Arr = { 3, 2, 1, 1, 4 };
int N = Arr.Length;
if(checkIndices(Arr, N))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is Contributed by Sachin
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the
// above approach
// Function that will tell whether
// such Indices present or Not.
function checkIndices($Arr, $N)
{
for ($i = 0; $i < $N - 1; $i++)
{
for ($j = $i + 1;
$j < $N; $j++)
{
// Checking 1st condition i.e
// whether Arr[i] equal to
// Arr[j] or not
if ($Arr[$i] != $Arr[$j])
{
// Checking 2nd condition i.e
// whether Arr[Arr[i]] equal to
// Arr[Arr[j]] or not.
if ($Arr[$Arr[$i] - 1] == $Arr[$Arr[$j] - 1])
return true;
}
}
}
return false;
}
// Driver Code
$Arr = array(3, 2, 1, 1, 4);
$N = sizeof($Arr);
// Calling function.
if(checkIndices($Arr, $N))
echo "Yes";
else
echo "No";
// This code is contributed
// by Akanksha Rai
?>
java 描述语言
<script>
// Javascript implementation of the above approach
// Function that will tell whether
// such Indices present or Not.
function checkIndices(Arr, N)
{
for (var i = 0; i < N - 1; i++) {
for (var j = i + 1; j < N; j++) {
// Checking 1st condition i.e whether
// Arr[i] equal to Arr[j] or not
if (Arr[i] != Arr[j]) {
// Checking 2nd condition i.e whether
// Arr[Arr[i]] equal to Arr[Arr[j]] or not.
if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])
return true;
}
}
}
return false;
}
// Driver Code
var Arr = [ 3, 2, 1, 1, 4 ];
var N = Arr.length;
// Calling function.
checkIndices(Arr, N) ? document.write( "Yes")
: document.write( "No");
</script>
Output:
Yes
时间复杂度: O(N 2 )
辅助空间: O(1)
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