难度: Medium

内容描述

给定n个整数的数组nums，其中n>1，返回一个数组输出，使得输出[i]等于nums[i]之外的所有nums元素的乘积。
Example:
Input:  [1,2,3,4]
Output: [24,12,8,6]
Note: Please solve it without division and in O(n).
Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

解题方案

思路 1
**- 时间复杂度: O(N)**- 空间复杂度: O(N)**

前缀积和后缀积, 懒得实现了

Follow up:

Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

思路 1
**- 时间复杂度: O(N)**- 空间复杂度: O(1)**

还是一样的思想，只不过不记录下来，而是采用边循环边更新的方式

beats 100.00%

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] res = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            res[i] = 1;
        }
        int left = 1;
        for (int i = 0; i < nums.length - 1; i++) {
            left *= nums[i];
            res[i+1] *= left;
        }
        int right = 1;
        for (int i = nums.length - 1; i > 0; i--) {
            right *= nums[i];
            res[i-1] *= right;
        }
        return res;
    }
}