标签：
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)
(medium)LeetCode 238.Product of Array Except Self
标签：
原文地址：http://www.cnblogs.com/mlz-2019/p/4702847.html
