【LeetCode】724. 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].
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/find-pivot-index 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
特别简单的题，可能是我心态不太好，没有读懂题，做的时候思路也不是很好，。。。。
直接放答案，这个思路一看就出来了
class Solution {
public:
int pivotIndex(vector<int>& nums) {
int sum = 0, leftsum = 0;
for (int x: nums) sum += x;
for (int i = 0; i < nums.size(); ++i) {
if (leftsum == sum - leftsum - nums[i]) return i;
leftsum += nums[i];
}
return -1;
}
};