375. Guess Number Higher or Lower II-优快云博客

本文链接：https://blog.youkuaiyun.com/smallplum123/article/details/71155589

本文探讨了一个猜数字游戏中的最优策略，目标是最小化猜中特定范围内随机数字所需的花费。通过动态规划方法解决了这一问题，并给出了详细的算法实现。

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1.Question

Total Accepted: 18911
Total Submissions: 53226
Difficulty: Medium
Contributor: LeetCode

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.

2.Code

class Solution {
public:
    int getMoneyAmount(int n) {
        int dp[n+1][n+1] = {0};
        for(int i = n; i > 0; i--)
            for(int j = i + 1; j <= n; j++)
            {
                dp[i][j] = i + dp[i+1][j];   //初始化dp[i][j],相当于k=i
                for(int k = i + 1; k < j; k++)  //k从i+1到j-1
                    dp[i][j] = min(dp[i][j], k + max(dp[i][k-1], dp[k+1][j]));
                dp[i][j] = min(dp[i][j], j + dp[i][j-1]);  // k = j
            }
        return dp[1][n];
    }
};

3.Note

1. 这个题的意思是，在1~n这个区间猜出一个数字，最少需要消耗多少钱，有个细节要注意，在k~k之间只有1个数，这时候是不需要消耗钱的。这个题用DP做，是一个min max的问题，和摔鸡蛋的问题有相似之处。对于区间[i, j], 猜 k 的话，那么需要消耗的钱数是k+max( dp[i][k-1], dp[k+1][j])，那么最小的消耗就是 min(k+max( dp[i][k-1], dp[k+1][j])), k = i,...,j。注意这里的边界处理，当k=i或者j的时候，写程序的时候是需要另外处理的，详情见代码。