Leetcode no. 238

最新推荐文章于 2019-12-07 10:14:56 发布

转载最新推荐文章于 2019-12-07 10:14:56 发布 · 268 阅读

leetcode 专栏收录该内容

114 篇文章

订阅专栏

本文介绍了一种在O(n)时间内求解数组中每个元素的乘积(排除自身)的方法，无需使用额外空间，同时提供了一个简洁的Java实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.)

Simple Java solution in O(n) without extra space

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