[LeetCode]Find Pivot_findpivot-优快云博客

本文介绍了一种算法，该算法能够在给定整数数组的情况下找到数组的支点索引。支点索引定义为数组中一个特殊的索引位置，在该位置左侧所有元素之和等于右侧所有元素之和。若不存在这样的索引则返回-1；若有多个这样的索引，则返回最左侧的索引。

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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.

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        if(nums.size()<3) return -1;
        int sum[nums.size()], max;
        for(int i = 0; i < nums.size(); i ++){
            i == 0?sum[i] = nums[i] : sum[i] = nums[i] + sum[i-1];
        }
        max  = sum[nums.size()-1];
        for(int i = 0; i < nums.size(); i++){
            if(max - sum[i] == sum[i] - nums[i]){
                return i;
            }
        }
        return -1;
    }
};