LeetCode 238. Product of Array Except Self（陣列元素的乘積）

阿新 • • 發佈：2019-01-21

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.)

思路：來回兩邊掃描，就可以保持兩個方向掃描時候，前面元素的連乘積。時間複雜度O(n)，空間複雜度O(n)。

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

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