本文主要是介绍LeetCode75——Day19，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
文章目录
- 一、题目
- 二、题解
一、题目
724. Find Pivot Index
Given an array of integers nums, calculate the pivot index of this array.
The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index’s right.
If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.
Return the leftmost pivot index. If no such index exists, return -1.
Example 1:
Input: nums = [1,7,3,6,5,6]
Output: 3
Explanation:
The pivot index is 3.
Left sum = nums[0] + nums[1] + nums[2] = 1 + 7 + 3 = 11
Right sum = nums[4] + nums[5] = 5 + 6 = 11
Example 2:
Input: nums = [1,2,3]
Output: -1
Explanation:
There is no index that satisfies the conditions in the problem statement.
Example 3:
Input: nums = [2,1,-1]
Output: 0
Explanation:
The pivot index is 0.
Left sum = 0 (no elements to the left of index 0)
Right sum = nums[1] + nums[2] = 1 + -1 = 0
Constraints:
1 <= nums.length <= 104
-1000 <= nums[i] <= 1000
二、题解
利用前缀和的思想做题
class Solution {
public:int pivotIndex(vector<int>& nums) {int n = nums.size();vector<int> prefixSum(n+1,0);for(int i = 1;i <= n;i++){prefixSum[i] = prefixSum[i-1] + nums[i-1];}for(int i = 0;i < n;i++){int left = prefixSum[i];int right = prefixSum[n] - prefixSum[i + 1];if(left == right) return i;}return -1;}
};
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