赞
踩
题目链接:https://leetcode.com/problems/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.)
Subscribe to see which companies asked this question
给出一个数组,要求计算一个新数组,数组里所有的元素都是除了自己以外的元素乘积。并且要求不许用除法。
《编程之美》上的一道原题。创建两个辅助数组,一个保存所有左边元素乘积的结果。一个保存所有右边元素乘积的结果。借助这两个数组,一次遍历就可以得到结果。
我的AC代码
- public class ProductofArrayExceptSelf {
-
- public static void main(String[] args) {
- int[] a = { 1, 2, 3, 4 };
- System.out.print(Arrays.toString((productExceptSelf(a))));
- }
-
- public static int[] productExceptSelf(int[] nums) {
- int len = nums.length;
- int[] r = new int[len];
-
- int[] left = new int[len];
- int[] right = new int[len];
- left[0] = nums[0];
- for (int i = 1; i < len; i++) {
- left[i] = left[i - 1] * nums[i];
- }
- right[len - 1] = nums[len - 1];
- for (int i = len - 2; i >= 0; i--) {
- right[i] = right[i + 1] * nums[i];
- }
-
- r[0] = right[1];
- r[len - 1] = left[len - 2];
- for (int i = 1; i < len - 1; i++) {
- r[i] = left[i - 1] * right[i + 1];
- }
- return r;
- }
- }
Copyright © 2003-2013 www.wpsshop.cn 版权所有,并保留所有权利。