LeetCode 375. Guess Number Higher or Lower II

原题链接在这里：https://leetcode.com/problems/guess-number-higher-or-lower-ii/

题目：

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?

题解：

是Guess Number Higher or Lower的进阶题. 

DP题. e.g. n = 6, 第一次猜3, 如果猜大了，那接下来要求的是从 [1, 2] 中猜到正确数字所需要花费的最少money, 记为x. 如果猜的小了，那么我们接下去要求的是从 [4, 6] 中猜到正确数字所需要花费的最少 money, 记为 y. 如果刚好猜中，则结束。

很显然，如果第一把猜 3, 那么要猜中数字预算需要花费的 money 为 3+ max(x, y). 因为需要的话费就是 做最坏的打算，尽最大努力，取最大值. 这是第一次取3的情况，还有5中情况. 取1~n所有情况下的最小值就是最小化将面临的可能的最大损失.

len是取出一段的长度. start是一段的开始.

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

Time Complexity: O(n^3). Space: O(n^2).

AC Java:

public class Solution {
    public int getMoneyAmount(int n) {
        int [][] dp = new int[n+1][n+1];
        for(int len = 2; len<=n; len++){
            for(int start = 1; start<=n-len+1; start++){
                int minPay = Integer.MAX_VALUE;
                for(int pivot = start; pivot < start+len-1; pivot++){
                    int temp = pivot + Math.max(dp[start][pivot-1], dp[pivot+1][start+len-1]);
                    minPay = Math.min(minPay, temp);
                }
                dp[start][start+len-1] = minPay;
            }
        }
        return dp[1][n];
    }
}

Reference: Minimax,

http://www.cnblogs.com/zichi/p/5701194.html