leetcode238

public class Solution {
    public int[] ProductExceptSelf(int[] nums) {
        int[] result = new int[nums.Length];
            for (int i = 0, tmp = 1; i < nums.Length; i++)
            {
                result[i] = tmp;
                tmp *= nums[i];
            }
            for (int i = nums.Length - 1, tmp = 1; i >= 0; 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