leetcode [238. 除自身以外数组的乘积]

(https://leetcode-cn.com/problems/product-of-array-except-self/)

感觉最近的每日一题都挺好玩的，静下心好好想想能体会到久违的思考的乐趣：）

方法一：就是开两个数组，每个数组存前缀积和后缀积，用到的时候直接拿出来相乘就行了

class Solution {
public:
    vector<int> ad1,ad2,ans;
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        ad1.resize(n+1);ad2.resize(n+1);
        ad1[0] = 1,ad2[n] = 1;
        for(int i = 1; i <= n; i++) ad1[i] = ad1[i-1] * nums[i-1];
        for(int i = n-1; i >= 0; i--) ad2[i] = ad2[i+1] * nums[i];
        for(int i = 0; i < n; i++){
            int cnt = ad1[i] * ad2[i+1];
            ans.push_back(cnt);
        }
        return ans;
    }
};

方法二：我用的是递归，虽然减少了数组空间，但增加了栈空间，但是再怎么说也是自己想出来的--

class Solution {
public:
    vector<int> ans;
    int N,right;
    void solve(int step,int left,vector<int>& nums){
        if(step == N-1){
            ans.push_back(left*right);
            return;
        }
        solve(step+1,left*nums[step],nums);
        right *= nums[step+1];
        ans.push_back(left*right);
    }
    vector<int> productExceptSelf(vector<int>& nums) {
        N = nums.size();
        right = 1;
        solve(0,1,nums);
        reverse(ans.begin(),ans.end());
        return ans;
    }
};

方法三：官方题解给出的空间O(1)的做法，看完竟然感觉有点...

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        vector<int> ans(n);
        ans[0] = 1;
        for (int i = 1; i < n; i++) ans[i] = ans[i-1]*nums[i-1];
        int r = 1;
        for (int i = n-1; i >= 0; i--){
            ans[i] *= r;
            r *= nums[i];
        }
        return ans;
    }
};