【Leetcode】Guess Number Higher or Lower II

题目链接：

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.Secondround:Youguess7,Itellyouthatit′shigher.Youpay7.

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.

思路：

关键在于把它转化为dp问题，O(n^3)的解法我总想不到。。/(ㄒoㄒ)/~~

设dp[i][j]为i~j要保证赢的最多代价。 假设猜k值，若不对，则问题转化为猜1~k-1,k+1~n两个子问题，且猜k的结果会告诉我们需要猜哪个子问题，所以猜k要保证赢 最多需要money=k+max{dp[i][k-1],dp[k+1,j]}，则在i~j范围内存在某策略是最优的，则是猜测某个k需要的最少的money，为min{money} k属于[i,j]。

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

算法：

    public int getMoneyAmount(int n) {
        int[][] dp = new int[n + 1][n + 1];
        for (int k = 1; k < n; k++) {
            for (int s = 1; s + k <= n; s++) {
                dp[s][s + k] = Integer.MAX_VALUE;
                for (int i = s; i < s + k; s++) {
                    dp[s][s + k] = Math.min(dp[s][s + k], i + Math.max(dp[s][i - 1], dp[i + 1][s + k]));
                }
            }
        }
        return dp[1][n];
    }