Product of Array Except Self

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.)

Solution1: run time complexity is O(n), space complexity is O(n).

res = [n2*n3*n4, n1*n3*n4, n1*n2*n4, n1*n2*n3], 相当于
a1 = [1, n1, n1*n2, n1*n2*n3]和
a2 = [n2*n3*n4, n3*n4, n4, 1]相乘

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] res = new int[nums.length];
        int[] a1 = new int[nums.length];
        int[] a2 = new int[nums.length];
        a1[0] = 1;
        a2[nums.length - 1] = 1;
        
        // scan from left to right -> a1 = [1, n1, n1*n2, n1*n2*n3]
        for (int i = 0; i < nums.length - 1; i++) {
            a1[i + 1] = nums[i] * a1[i];
        }
        
        // scan from right to left -> a2 = [n2*n3*n4, n3*n4, n4, 1]
        for (int i = nums.length - 1; i > 0; i--) {
            a2[i - 1] = nums[i] * a2[i];
        }
        
        // multiply -> res = [n2*n3*n4, n1*n3*n4, n1*n2*n4, n1*n2*n3]
        for (int i = 0; i < nums.length; i++) {
            res[i] = a1[i] * a2[i];
        }
        
        return res;
    }
}

Solution2: run time complexity is O(n), constant space.

用一个temp来保存每次的结果。

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