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.)
解法:output[p]=pre[p]*after[p]
代码如下:
public class Solution { public int[] productExceptSelf(int[] nums) { int len=nums.length; int[] output=new int[len]; int[] pre=new int[len]; int[]after=new int[len]; for(int i=0;i<len;i++) pre[i]=1; for(int j=0;j<len;j++) after[j]=1; for(int m=1;m<len;m++){ pre[m]=pre[m-1]*nums[m-1]; } for(int n=len-2;n>=0;n--){ after[n]=after[n+1]*nums[n+1]; } for(int p=0;p<len;p++){ output[p]=pre[p]*after[p]; } return output; } }
运行结果:时间复杂度O(n)
时间: 2024-11-03 21:43:05