LeetCode724. Find Pivot Index-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_33256893/article/details/78826545

本文介绍了一种高效的方法来解决LeetCode上寻找数组中心索引的问题，即找到数组中一个特殊的索引位置，该位置左侧所有元素的总和等于右侧所有元素的总和。通过先计算整个数组的总和，再从左到右遍历数组，可以快速确定是否存在这样的中心索引。

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原题：https://leetcode.com/problems/find-pivot-index/description/

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].

求一个pivot是的pivot左右两边之和相等，实质就是求某一边和等于除去pivot的所有数的和的一半。
首先遍历一遍数组求得所有数之和，然后再从左遍历一次，每次把当前左边所有数的和与先前求得的所有数的和减去当前pivot所指的数再减半后的值对比（if (leftsum == (sum-nums[i])/2)），如果相等则表明当前左右两边相等，如果没有找到这个数，则返回-1.

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        double sum = 0;
        for (int i = 0; i < nums.size(); i++) {
            sum += nums[i];
        }
        double leftsum = 0;
        for (int i = 0; i < nums.size(); i++) {
            if (leftsum == (sum-nums[i])/2) {
                return i;
            } else {
                leftsum += nums[i];
            }
        }
        return -1;
    }
};