技術標籤：ArrayLeetcode

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;
    }
}