Product of Array Except Self
问题描述
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.)
解决思路
1. 使用辅助空间
为方便理解,使用两个与输入数组同等大小的辅助数组,
首先,从前到后扫描数组,一个数组[i]存储的是除了当前元素外的所有前面元素的乘积;
然后,从后到前扫描数组,另一个数组[i]存储的是除了当前元素外的所有后面元素的乘积。
最后将这两个数组的对应位置的元素相乘即可。
优化后可以省略一个数组和一遍扫描。
程序
public class Solution { public int[] productExceptSelf(int[] nums) { if (nums == null) { return null; } int len = nums.length; int[] res = new int[len]; int[] tmp = new int[len]; tmp[0] = 1; res[0] = 1; for (int i = 1; i < len; i++) { tmp[i] = tmp[i - 1] * nums[i - 1]; res[i] = tmp[i]; } tmp[len - 1] = 1; for (int i = len - 2; i >= 0; i--) { tmp[i] = tmp[i + 1] * nums[i + 1]; res[i] *= tmp[i]; } return res; } }