724. Find Pivot Index （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].

這道題用python做速度很慢。

剛開始一直超時，因為每一次迭代都分別前後求和再比較：sum(left) == sum(right)，這樣的後果就是超級耗時！！！

後來看了discussion裡的解法，要記錄sum和left，right = sum - left - nums[i],或者用 left* 2 + nums[i] == sum來判斷

class Solution:
    def pivotIndex(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        length = len(nums)
        sm = sum(nums)
        l = 0
        for i in range(length):
            if(l*2 == sm - nums[i]): return i
            l += nums[i]
        return -1