【Leetcode】238-Product of Array Except Self【Java实现】【Medium】【two way traverse】

复杂度要求O(n)，不代表只能遍历一遍，可以遍历2遍、3遍。。。。

product--->名词：积

stem:

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.)

code:

public class Solution {/*by qr2015-11-7two way traverseerror:[0,4,0]*/    public int[] productExceptSelf(int[] nums) {        int len=nums.length;        int [] res=new int[len];                res[0]=1;        for(int i=1;i<len;i++){            res[i]=res[i-1]*nums[i-1];        }                int temp=1; //要乘以temp，而不是nums[i+1]        for(int i=len-2;i>=0;i--){            temp*=nums[i+1];            res[i]=res[i]*temp;        }                return res;    }}

采用了two way traverse，因为最后的结果就可以分为两个部分，左部分和右部分，所以分为两个方向的乘积。