題目：

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 
equalto 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].

解釋：
找到陣列nums中的pivot的位置，使得其左邊的元素的總和等於其右邊的元素的總和，如果不存在則返回-1

class Solution(object):
    def pivotIndex(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        full_sum=sum(nums)
        left=0
        right=0
        len_nums=len(nums)
        for i in xrange(len_nums):
            if i>0:
                left+=nums[i-1]
            right=full_sum-left-nums[i]
            if left==right:
                return i
        return -1

c++程式碼：

class Solution {
public:
    int pivotIndex(vector<int>& nums) {
        int left_sum=0;
        int right_sum=0;
        int n=nums.size();
        int full_sum=accumulate(nums.begin(),nums.end(),0);
        for (int i=0;i<n;i++)
        {
            if (i>0)
                left_sum+=nums[i-1];
            right_sum=full_sum-left_sum-nums[i];
            if (left_sum==right_sum)
                return i;
        }
        return -1;
    }
};

總結：
儘量不要每次都暴力計算，要充分利用歷史求和結果。