leetcode238

public class Solution {
    public int[] ProductExceptSelf(int[] nums) {
        int[] result = new int[nums.Length];
            for (int i = , tmp = ; i < nums.Length; i++)
            {
                result[i] = tmp;
                tmp *= nums[i];
            }
            for (int i = nums.Length - , tmp = ; i >= ; i--)
            {
                result[i] *= tmp;
                tmp *= nums[i];
            }
            return result;
    }
}

https://leetcode.com/problems/product-of-array-except-self/#/description

在第一个循环中，记录当前的索引左侧的数字乘积，和当前的索引右侧的数字乘积。然后两部分乘积相乘，这样的思路比较简单，但是时间复杂度是O(n*n)。

本题的解法时间复杂度更低，但是代码不容易理解。

补充一个python的实现：

 class Solution:
     def productExceptSelf(self, nums: 'List[int]') -> 'List[int]':
         n = len(nums)
         front = [1] * n
         front[0] = nums[0]

         back = [1] * n
         back[n-1] = nums[n-1]

         for i in range(1,n):
             front[i] = front[i-1] * nums[i]

         for i in range(n-2,0,-1):
             back[i] = back[i+1] * nums[i]

         l = list()
         for i in range(n):
             if i == 0:
                 f = 1
                 b = back[i+1]
                 l.append(f*b)
             elif i == n-1:
                 f = front[i-1]
                 b = 1
                 l.append(f*b)
             else:
                 f = front[i-1]
                 b = back[i+1]
                 l.append(f*b)
         return l