【Leetcode】Product of Array Except Self

题目链接：https://leetcode.com/problems/product-of-array-except-self/

题目：

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

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

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

思路：

1、规避除法。用两个数组描述元素左、右边乘积，结果就是该元素左右乘积之积。空间复杂度为O(n)

2、在上述思路基础上减少数组的使用，用result代替右边的乘积，再用一个常数表示元素左边乘积。

算法1：

public int[] productExceptSelf(int[] nums) {int right[] = new int[nums.length]; //元素右边乘积int left[] = new int[nums.length];//int result[] = new int[nums.length];for (int i = nums.length - 1; i >= 0; i--) {if (i == nums.length - 1) {right[i] = 1;} else {right[i] = right[i + 1] * nums[i + 1];}}for (int i = 0; i < nums.length; i++) {if (i == 0) {left[i] = 1;} else {left[i] = left[i - 1] * nums[i - 1];}}for (int i = 0; i < nums.length; i++) {result[i] = right[i] * left[i];}return result;}

算法2：

public int[] productExceptSelf(int[] nums) {int result[] = new int[nums.length];for (int i = nums.length - 1; i >= 0; i--) { //元素右边乘积if (i == nums.length - 1) {result[i] = 1;} else {result[i] = result[i + 1] * nums[i + 1];}}int left = 0;//左边乘积for (int i = 0; i < nums.length; i++) {if (i == 0) {left = 1;} else {left= left * nums[i - 1];result[i] = result[i] * left;}}return result;}