238. Product of Array Except Self

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 numsexcept 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.)

class Solution { //O(n)
    public int[] productExceptSelf(int[] nums) {
        int[] res = new int[nums.length];
        int[] left = new int[nums.length];
        int[] right = new int[nums.length];
        left[0] = 1;
        right[nums.length-1] = 1;
        for(int i = 1; i < nums.length; i++){
            left[i] = left[i-1] * nums[i-1];
        }
        for(int i = nums.length-2; i >= 0; i--){
            right[i] = right[i+1] * nums[i+1];
        }
        for(int i = 0; i < nums.length; i++){
            res[i] = left[i] * right[i];
        }
        return res;
    }
}

我们以一个4个元素的数组为例，nums=[a1,a2,a3,a4]，要想在O(n)的时间内输出结果，比较好的解决方法是提前构造好两个数组：

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

然后两个数组一一对应相乘，即可得到最终结果 [a2*a3*a4, a1*a3*a4, a1*a2*a4, a1*a2*a3]。

不过，上述方法的空间复杂度为O(n)，可以进一步优化成常数空间，即用一个整数代替第二个数组。'

public class Solution {//O(1)
    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;
    }
}