375. Guess Number Higher or Lower II

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最新推荐文章于 2024-03-18 09:44:26 发布

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本文介绍了一个猜数字游戏的最优策略，目标是最小化最坏情况下的支付总额。通过动态规划方法，采用区间型DP思想，从短区间推导长区间，最终求解出在给定范围内必胜所需的最小资金。

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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.

摘自

https://discuss.leetcode.com/topic/51358/java-dp-solution

https://discuss.leetcode.com/category/495/guess-number-higher-or-lower-ii

Definition of dp[i][j]: minimum number of money to guarantee win for subproblem [i, j].

Target: dp[1][n]

Corner case: dp[i][i] = 0 (because the only element must be correct)

Equation: we can choose k (i<=k<=j) as our guess, and pay price k. After our guess, the problem is divided into two subproblems. Notice we do not need to pay the money for both subproblems. We only need to pay the worst case (because the system will tell us which side we should go) to guarantee win. So dp[i][j] = min (i<=k<=j) { k + max(dp[i][k-1], dp[k+1][j]) }

public class Solution {
    public int getMoneyAmount(int n) {
        if (n == 1) {
            return 0;
        }
        int[][] dp = new int[n + 1][n + 1];
        for (int jminusi = 1; jminusi < n; jminusi++) {
            for (int i = 0; i + jminusi <= n; i++) {
                int j = i + jminusi;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i; k <= j; k++) {
                    dp[i][j] = Math.min(dp[i][j],
                                        k + Math.max(k - 1 >= i ? dp[i][k - 1] : 0,
                                                     j >= k + 1 ? dp[k + 1][j] : 0));
                }
            }
        }
        return dp[1][n];
    }
}

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区间型DP先求出短的区间，由短的区间推出长的区间。

我们需要按照一定的顺序计算dp的值，由于计算dp[i][j]时我们需要计算dp[i][j-1],dp[i+1][j]，所以我们按照j-i递增的顺序计算，即l区间长度len由短到长的顺序计算。

public static int getMoneyAmount(int n)
	{
        if (n == 1)
            return 0;
        
        int[][] dp = new int[n + 1][n + 1];
        
        for (int len = 1; len < n; len++)
            for (int i = 0; i + len <= n; i++)
            {
                int j = i + len;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i; k <= j; k++) 
                    dp[i][j] = Math.min(dp[i][j],k + Math.max(k-1>=i?dp[i][k - 1]:0,k+1<=j?dp[k + 1][j]:0));
            }
        
        return dp[1][n];
    }

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记忆优化搜索

public class Solution {
    public int getMoneyAmount(int n) {
        int[][] table = new int[n+1][n+1];
        return DP(table, 1, n);
    }
    
   public int DP(int[][] t, int s, int e){
        if(s >= e) return 0;
        if(t[s][e] != 0) return t[s][e];
        int res = Integer.MAX_VALUE;
        for(int x=s; x<=e; x++){
            int tmp = x + Math.max(DP(t, s, x-1), DP(t, x+1, e));
            res = Math.min(res, tmp);
        }
        t[s][e] = res;
        return res;
    }
}