724.Find Pivot Index

724.Find Pivot Index

难度:Easy

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].

找到一条平分轴,使得左右两边和相等。可以先求出总和,然后从左开始递增,每次比较左边的和与总和减去轴上的值的一半是否相等,相等则返回轴的索引。

class Solution {
public:
int pivotIndex(vector<int>& nums) {
if(nums.empty()) return -1;
int sum= accumulate(nums.begin(),nums.end(),0);
int leftSum=0;
for(int index=0;index<nums.size();index++)
{
int tmp=sum-nums[index];
if(tmp%2 ==0 && leftSum ==tmp/2)
return index;
leftSum += nums[index];
}
return -1;
}
};

执行用时 :28 ms, 在所有 C++ 提交中击败了85.29%的用户 内存消耗 :9.8 MB, 在所有 C++ 提交中击败了89.77%的用户