在这个问题中，我们给出了一个大小为N的数组arr[]。我们的任务是在数组中可能的移动后找到左指针的索引。
我们有两个指针，一个是左指针，另一个是右指针。
左指针从索引0开始，值递增。
右指针从索引(n-1)开始，值递减。
如果左指针的和小于右指针的和，则指针的值增加，否则指针的值减少。并且和的值会更新。
让我们通过一个例子来理解这个问题，
Input : arr[] = {5, 6, 3, 7, 9, 4}
Output : 2Explanation −
的中文翻译为：Explanation −
leftPointer = 0 -> sum = 5, rightPointer = 5 -> sum = 4. Move rightPointer leftPointer = 0 -> sum = 5, rightPointer = 4 -> sum = 13. Move leftPointer leftPointer = 1 -> sum = 11, rightPointer = 4 -> sum = 13. Move leftPointer leftPointer = 2 -> sum = 14, rightPointer = 4 -> sum = 13. Move rightPointer leftPointer = 2 -> sum = 14, rightPointer = 3 -> sum = 20. Move rightPointer Position of the left pointer is 2.
通过根据和的大小移动左指针和右指针来解决问题。然后检查左指针是否比右指针大1。
程序示例，说明我们解决方案的工作原理
#include <iostream>
using namespace std;
int findIndexLeftPointer(int arr[], int n) {
if(n == 1)
return 0;
int leftPointer = 0,rightPointer = n-1,leftPointerSum = arr[0], rightPointerSum = arr[n-1];
while (rightPointer > leftPointer + 1) {
if (leftPointerSum < rightPointerSum) {
leftPointer++;
leftPointerSum += arr[leftPointer];
}
else if (leftPointerSum > rightPointerSum) {
rightPointer--;
rightPointerSum += arr[rightPointer];
}
else {
break;
}
}
return leftPointer;
}
int main() {
int arr[] = { 5, 6, 3, 7, 9, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"The index of left pointer after moving is "<<findIndexLeftPointer(arr, n);
return 0;
}The index of left pointer after moving is 2
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