## 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?
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：

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：

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

## Python代码：

``````class Solution(object):
def getMoneyAmount(self, n):
"""
:type n: int
:rtype: int
"""
dp = [ * (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[n]
``````

## Python代码：

``````class Solution(object):
def getMoneyAmount(self, n):
"""
:type n: int
:rtype: int
"""
dp = [ * (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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