724. Find Pivot Index*
https://leetcode.com/problems/find-pivot-index/description/
题目描述
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].
解题思路
首先求出整个数组的和 sum, 然后依次遍历每一个索引 i, 判断左边的和 leftsum 是否等于 sum - nums[i] - leftsum (即右边的和, 注意不包括 nums[i]), 如果不相等, 那么 leftsum += nums[i].
主要注意下面这种情况:
nums = [1, -1, 1]
index = 0
对于 nums[0] 左边没有值, 那么就认为是左边的和为 0.
C++ 实现 1
class Solution {
public:
int pivotIndex(vector<int>& nums) {
if (nums.size() < 3) return -1;
int sum = std::accumulate(nums.begin(), nums.end(), 0);
int left_val = 0;
for (int i = 0; i < nums.size(); ++i) {
int right_val = sum - left_val - nums[i];
if (right_val == left_val) return i;
left_val += nums[i];
}
return -1;
}
};
