leetcode238- Product of Array Except Self- medium

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.)

 

两个数组，一个存从最左边直到前一个数字的乘积，一个从最右边直到后一个数字的乘积。最后的答案就是这两个数组对应位置相乘。

空间优化，可以直接左边一次把第一个数组乘到最终结果里，右边扫一次把第二个数组乘到最终结果里。

 

1.空间优化后

class Solution {
    public int[] productExceptSelf(int[] nums) {
        
        if (nums == null || nums.length == 0) {
            return new int[0];
        }
        
        int product = 1;
        int[] result = new int[nums.length];
        
        for (int i = 0; i < nums.length; i++) {
            result[i] = product;
            product *= nums[i];
        }
        
        product = 1;
        for (int i = nums.length - 1; i >= 0; i--) {
            result[i] *= product;
            product *= nums[i];
        }
        
        return result;
    }
}

 

2.空间优化前

class Solution {
    public int[] productExceptSelf(int[] nums) {
        
        if (nums == null || nums.length == 0) {
            return new int[0];
        }
        
        int[] forward = new int[nums.length];
        int[] backward = new int[nums.length];
        int[] result = new int[nums.length];
        
        forward[0] = 1;
        for (int i = 1; i < nums.length; i++) {
            forward[i] = forward[i - 1] * nums[i - 1];
        }
        backward[nums.length - 1] = 1;
        for (int i = nums.length - 2; i >= 0; i--) {
            backward[i] = backward[i + 1] * nums[i + 1];
        }
        
        for (int i = 0; i < nums.length; i++) {
            result[i] = forward[i] * backward[i];
        }
        return result;
    }
}