Haiyang Yu's Blog
0%

Leetcode 724 Find Pivot Index

发表于 更新于 分类于 Valine：
本文字数： 1.3k 阅读时长 ≈ 1 分钟

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 all the numbers to the left of the index is equal to the sum of all 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:

1
2
3
4
5
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:

1
2
3
4
Input: nums = [1,2,3]
Output: -1
Explanation:
There is no index that satisfies the conditions in the problem statement.

Constraints:

  • The length of nums will be in the range [0, 10000].
  • Each element nums[i] will be an integer in the range [-1000, 1000].

遍历

维护两个值leftSum和rightSum, 一开始假设i = 0. i左边没有元素,所以leftsum = 0; i右边的和为accumulate(nums.begin() + 1, nums.end(), 0). 依次判断leftsum和rightsum是否相等. 相等则直接返回i, 不相等i需要向后移动一位, 这时不用重新计算新的leftsum和rightsum, 只需要在i移动到i+1之前, 将leftsum += nums[i] 和rightsum -= nums[i+1]即可求的新的leftsum,rightsum.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        if(nums.size() == 0)
            return -1;
        if(nums.size() == 1)
            return 0;
        int leftSum = 0;
        int rightSum = accumulate(nums.begin() + 1, nums.end(), 0);
        int i = 0;
        while(leftSum != rightSum)
        {
            if(i == nums.size() - 1)
                return -1;
            leftSum += nums[i];
            rightSum -= nums[i + 1];
            ++i;
        }
        return i;
    }
};