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

複雜度要求O(n)，不代表只能遍歷一遍，可以遍歷2遍、3遍。。。。

product--->名詞：積

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 qr
2015-11-7
two way traverse
error:[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;
    }
}

採用了two way traverse，因爲最後的結果就可以分爲兩個部分，左部分和右部分，所以分爲兩個方向的乘積。