238. Product of Array Except Self- 非自身数组的乘积

最新推荐文章于 2021-07-22 23:09:09 发布

原创最新推荐文章于 2021-07-22 23:09:09 发布 · 512 阅读

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本文提供了一种解决LeetCode第238题“除自身以外数组的乘积”的方法，该方法使用O(n)的时间复杂度，并考虑了数组中包含0的情况。

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238. 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.)

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int count = 1;
        int count1 = 1;
        HashSet<Integer> set= new HashSet<Integer>();
        int[] result= new int[nums.length];
        for(int i = 0; i< nums.length; i++){
            count*=nums[i];
            if(nums[i]==0&&set.add(nums[i])) {
            	set.add(nums[i]);
            	continue;
            }
            count1*=nums[i];
        }
        for(int i = 0; i< nums.length; i++){
            if(nums[i]!=0) result[i]=count/nums[i];
            if(nums[i]==0) result[i]= count1;
        }
        return result;
    }
}

感觉想的有点麻烦，应该有简单方法。TO be continue……

原文链接：https://leetcode.com/problems/product-of-array-except-self/