题目描述：

LeetCode 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.

Hint:

1. The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario.
2. Take a small example (n = 3). What do you end up paying in the worst case?
3. Check out this article if you're still stuck.
4. The purely recursive implementation of minimax would be worthless for even a small n. You MUST use dynamic programming.
5. As a follow-up, how would you modify your code to solve the problem of minimizing the expected loss, instead of the worst-case loss?

题目大意：

我们玩猜数字游戏。游戏如下：

我从1到n中选择一个数字。你需要猜我选的是哪个数字。

每一次你猜错，我都会告诉你数字是高了还是低了。

然而，当你选择某个数字x并且猜错，你需要支付$x。当你猜到我选的数字时胜出。

测试用例如题目描述。

给定n ≥ 1，计算你需要多少钱才可以确保赢得游戏。

提示：

1. 游戏的最佳策略是最小化你面临的最大可能损失。另一种策略是最小化期望损失。在这里，我们关注第一种策略。
2. 以一个小的输入为例（n = 3）。最坏情况下你会支付多少金额？
3. 如果还是一筹莫展，可以参考这篇文章。
4. 单纯的minimax递归实现即使对于很小的n都是非常耗时的。你必须使用动态规划。
5. 作为思考题，你怎样修改代码来实现最小化的期望损失，而不是最小化的最坏损失？

解题思路：

动态规划（Dynamic Programming）

参考：https://discuss.leetcode.com/topic/51356/two-python-solutions

状态转移方程：

dp[i][j] = min(k + max(dp[i][k - 1], dp[k + 1][j]))

其中dp[i][j]表示猜出范围[i, j]的数字需要花费的最少金额。

Python代码：

class Solution(object):
    def getMoneyAmount(self, n):
        """
        :type n: int
        :rtype: int
        """
        dp = [[0] * (n+1) for _ in range(n+1)]
        for gap in range(1, n):
            for lo in range(1, n+1-gap):
                hi = lo + gap
                dp[lo][hi] = min(x + max(dp[lo][x-1], dp[x+1][hi])
                                   for x in range(lo, hi))
        return dp[1][n]

记忆化搜索：

自顶向下（Top-down Approach）求解，采用辅助数组dp记录已经计算过的结果，减少重复计算。

Python代码：

class Solution(object):
    def getMoneyAmount(self, n):
        """
        :type n: int
        :rtype: int
        """
        dp = [[0] * (n+1) for _ in range(n+1)]
        def solve(lo, hi):
            if lo < hi and dp[lo][hi] == 0:
                dp[lo][hi] = min(x + max(solve(lo,x-1), solve(x+1,hi))
                               for x in range(lo, hi))
            return dp[lo][hi]
        return solve(1, n)

 

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