[LeetCode] (medium) 375. Guess Number Higher or Lower II

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I'll tell you whether the number I picked is higher or lower.

However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.

Example:

n = 10, I pick 8.

First round:  You guess 5, I tell you that it's higher. You pay $5.
Second round: You guess 7, I tell you that it's higher. You pay $7.
Third round:  You guess 9, I tell you that it's lower. You pay $9.

Game over. 8 is the number I picked.

You end up paying $5 + $7 + $9 = $21.

Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.

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分治法，参考： https://www.cnblogs.com/zichi/p/5701194.html

记忆式DFS：

class Solution {
public:
    
    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n+1, vector<int>(n+1, -1));
        return dfs(1, n, dp);
    }
    
    int dfs(int lo, int hi, vector<vector<int>> &dp){
        if(lo > hi) return 0;
        if(lo == hi) return 0;
        if(dp[lo][hi] != -1) return dp[lo][hi];
        int tem = INT_MAX;
        for(int i = lo; i <= hi; ++i){
            tem = min(tem, i+max(dfs(lo, i-1, dp), dfs(i+1, hi, dp)));
        }
        dp[lo][hi] = tem;
        return tem;
    }
};

真正的DP：

class Solution {
public:
    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n+1, vector<int>(n+1, 0));
        for(int hi = 2; hi <= n; ++hi){
            for(int lo = hi-1; lo > 0; --lo){
                int tem = INT_MAX;
                for(int cur = lo+1; cur < hi; ++cur){
                    tem = min(tem, cur+max(dp[cur+1][hi], dp[lo][cur-1]));
                }
                dp[lo][hi] = (lo == hi-1) ? lo : tem;
            }
        }
        return dp[1][n];
    }
};

正确性可用数学归纳法证明，所以lo要反向遍历，这个思想值得学习