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 $$ n \geq 1 $$ , find out how much money you need to have to guarantee a win.

注意:题目给出的要求是保证一次赢必须的钱的数量 这就意味着[i, j]假设正确的数字是在[i, j]之间，为了保证赢，那么玩家必须承受最大的损失。

For each number x in range[i~j]
we do: result_when_pick_x = x + max{DP([i~x-1]), DP([x+1, j])}
--> // the max means whenever you choose a number, the feedback is always bad and therefore leads you to a worse branch.
then we get DP([i~j]) = min{xi, ... ,xj}
--> // this min makes sure that you are minimizing your cost.

递归的方式:

动态规划的方式

class Solution {
public:
    int getMoneyAmount(int n) {

        //dp[i][j]represent[i, j]
        vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
        for (int j = 1; j <= n; ++j){
            for (int i = j - 1;  i >= 1; --i){
                dp[i][j] = INT_MAX;
                for (int k = i; k <= j; ++k){
                    if(k == i)  
                        dp[i][j] = min(dp[i][j], k + dp[i+1][j]);
                    else if(k == j) 
                        dp[i][j] = min(dp[i][j-1] + j, dp[i][j]);
                    else{
                        dp[i][j] = min(dp[i][j],k + max(dp[i][k - 1], dp[k + 1][j]));
                    }
                }
            }
        }
        return dp[1][n];
    }
};