Product of Array Except Self

数组除了自己剩余数的乘积

题目

Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

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.)

解析重点

1.一开始我想着计算数组所有数的乘积，然后一个循环除以每个数自身即可，但是这里不让用除法。
2.那么我们怎么得到结果呢？a[i]=a[1]…a[i-1]a[i+1]*…a[n],根据上式，我们可以以a[i]为界限，把计算分为两部分，左边和右边乘积。这样我们只需要把数组
循环两次，第一次计算左边的乘积，第二次计算右边的乘积，合起来即为结果。

java代码

class Solution {
    public int[] productExceptSelf(int[] nums) {
    int n = nums.length;
    int[] res = new int[n];
    res[0] = 1;
	//计算左边乘积
    for (int i = 1; i < n; i++) {
        res[i] = res[i - 1] * nums[i - 1];
    }
    int right = 1;
    for (int i = n - 1; i >= 0; i--) {
        res[i] *= right;//将右边乘积合并到左边
        right *= nums[i];//计算右边乘积
    }
    return res;
    }
}