[LeetCode] Find Pivot Index

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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.

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].

求一个数组中某索引前的元素和等于索引后（包含索引）的元素和。感觉是一个动态规划题。

首先建立一个sums数组，里面存放元素逐个累加的和。例如

index =      0, 1, 2, 3, 4, 5

nums = [    1, 7, 3, 6, 5, 6]

index =  0, 1, 2,  3,   4,   5,   6

sums = [0, 1, 8, 11, 17, 22, 28]

sums[i]表示当前i索引之前所有值的和。

sums[n] - sums[i + 1]表示当前索引之后所有值的和。

循环遍历sums数组，当遍历到第i个时，判断当前sums[i]是否等于数组和即sums[n]与sums[i + 1]之差。如果相等返回索引i即可。

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int n = nums.size();
        vector<int> sums(n + 1, 0);
        for (int i = 1; i < sums.size(); i++) {
            sums[i] = sums[i - 1] + nums[i - 1];
        }
        for (int i = 0; i < sums.size() - 1; i++) {
            if (sums[i] == sums[n] - sums[i + 1]) {
                return i;
            }
        }
        return -1;
    }
};
// 35 ms