Given an array of integers nums, write a method that returns the "pivot" index of this array.

We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index.

If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.

Example 1:

Input: 
nums = [1, 7, 3, 6, 5, 6]
Output: 3
Explanation: 
The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
Also, 3 is the first index where this occurs.

 

Example 2:

Input: 
nums = [1, 2, 3]
Output: -1
Explanation: 
There is no index that satisfies the conditions in the problem statement.

 

Note:

  • The length of nums will be in the range [0, 10000].
  • Each element nums[i] will be an integer in the range [-1000, 1000].

题意：寻找数组的中心索引的下标，题目中对中心索引的定义是这个数左右两边的数字的和相等，如果找不到这样的中心索引，返回-1。

没想到其它什么方法，就遍历了数组，然后分别求它左右两边数字的和，看是否相等，如果相等则返回下标，遍历完数组还没有return，就返回-1。

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int len=nums.size();
        int suml=0,sumr=0;
        int i;
        for(i=0;i<len;i++){
            for(int j=0;j<i;j++){
                suml+=nums[j];
            }
            for(int k=i+1;k<len;k++){
                sumr+=nums[k];
            }
            if(suml==sumr){
                return i;
                break;
            }
            suml=0;
            sumr=0;

        }
        if(i==len) return -1;
        
    }
};