leetcode 238. Product of Array Except Self (乱搞)

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.)

 先来个O(n)空间的无脑解法。。。一个前缀积一个后缀积就好了。。。

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        vector<int>res;
        vector<int>pre,suf;
        int siz = nums.size();
        if (siz==0) return res;
        for ( int i = 0 ; i < siz ; i++)
        {
            int tmp;
            if (i==0) tmp = nums[0];
            else tmp = pre[i-1]*nums[i];
            pre.push_back(tmp);
        }
        int cnt = 0 ;
        for ( int i = siz -1 ; i>= 0 ; i--)
        {
            int tmp;
            if (i==siz-1) tmp = nums[siz-1];
            else tmp = suf[cnt++]*nums[i];
            suf.push_back(tmp);
        }
      //  for ( int i = 0 ; i < siz ; i++) printf("%d ",pre[i]); puts("");
    //    for ( int i = 0 ; i < siz ; i++) printf("%d ",suf[i]);puts("");
        reverse(suf.begin(),suf.end());
        for ( int i = 0 ; i < siz;  i++)
        {
            int l = i-1;
            int r = i+1;
            int x,y;
            if (l<0) x = 1;
            else x = pre[l];
            if (r>=siz) y = 1;
            else y = suf[r];
      //      cout<<"x:"<<x<<" y:"<<y<<endl;
            res.push_back(x*y);
        }
        return res;
        
    }
};

常数空间的做法。。。想了半天没有想法。。。

看了solution...发现就是借用res数组搞事情。。。。每个位置的答案扫完两遍得到。。。

感觉。。。非常的。。。无聊啊。。。。真心没意思吧，还以为是什么巧妙的做法呢。。。