# Product of Array Except Self

### 描述​

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].

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.)

### 分析​

• [1, a1, a1*a2, a1*a2*a3]
• [a2*a3*a4, a3*a4, a4, 1]

### 代码 1 O(n)空间​

// Product of Array Except Self// Time Complexity: O(n), Space Complexity: O(n)public class Solution {    public int[] productExceptSelf(int[] nums) {        final int[] result = new int[nums.length];        final int[] left = new int[nums.length];        final int[] right = new int[nums.length];        left[0] = 1;        right[nums.length - 1] = 1;        for (int i = 1; i < nums.length; ++i) {            left[i] = nums[i - 1] * left[i - 1];        }        for (int i = nums.length - 2; i >= 0; --i) {            right[i] = nums[i + 1] * right[i + 1];        }        for (int i = 0; i < nums.length; ++i) {            result[i] = left[i] * right[i];        }        return result;    }}

### 代码 2 O(1)空间​

// Product of Array Except Self// Time Complexity: O(n), Space Complexity: O(1)public class Solution {    public int[] productExceptSelf(int[] nums) {        final int[] left = new int[nums.length];        left[0] = 1;        for (int i = 1; i < nums.length; ++i) {            left[i] = nums[i - 1] * left[i - 1];        }        int right = 1;        for (int i = nums.length - 1; i >= 0; --i) {            left[i] *= right;            right *= nums[i];        }        return left;    }}