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 : 2
explanation −
的中文翻译为：explanation −
leftpointer = 0 -> sum = 5, rightpointer = 5 -> sum = 4. move rightpointerleftpointer = 0 -> sum = 5, rightpointer = 4 -> sum = 13. move leftpointerleftpointer = 1 -> sum = 11, rightpointer = 4 -> sum = 13. move leftpointerleftpointer = 2 -> sum = 14, rightpointer = 4 -> sum = 13. move rightpointerleftpointer = 2 -> sum = 14, rightpointer = 3 -> sum = 20. move rightpointerposition of the left pointer is 2.
解决方案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。
示例程序示例，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;}
输出the index of left pointer after moving is 2
以上就是在c++中，找到数组中可能移动后左指针的索引的详细内容。