2020-12-12 238. Product of Array Except Self

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本文介绍了一种算法问题，即给定一个整数数组，返回一个输出数组，其中输出数组的每个元素等于输入数组中除当前索引外所有元素的乘积。提供了两种解决方案：一种使用O(n)时间和O(n)额外空间；另一种使用O(n)时间复杂度但仅需O(1)额外空间。

238. Product of Array Except Self

Difficulty: Medium

Related Topics: Array

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 nums except nums[i].

Example:

Input:  [1,2,3,4]
Output: [24,12,8,6]

Constraint: It’s guaranteed that the product of the elements of any prefix or suffix of the array (including the whole array) fits in a 32 bit integer.

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.)

Solution 1 using O(n) time and O(n) space

Keys: This solution makes use of divide and conquer, we have to seperate our solution into different pieces and combine them together.

Language: Java

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

Solution 2 using O(n) time and O(1) space

Keys: This solution is quite similar, though it combines one of the for loops with the result for loop.

Language: Java

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