LeetCode 375. Guess Number Higher or Lower II

原题网址：https://leetcode.com/problems/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 firstscenario.
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?

方法：MiniMax问题，一定要搞清楚min函数和max函数。

public class Solution {
    private int[][] amounts;
    private int min(int from, int to) {
        if (from >= to) return 0;
        if (amounts[from][to] > 0) return amounts[from][to];
        int min = Integer.MAX_VALUE;
        for(int i = from; i <= to; i++) {
            min = Math.min(min, max(from, to, i));
        }
        amounts[from][to] = min;
        return min;
    }
    private int max(int from, int to, int guess) {
        if (from >= to) return 0;
        int max = guess + Math.max(min(from, guess - 1), min(guess + 1, to));
        return max;
    }
    public int getMoneyAmount(int n) {
        amounts = new int[n + 1][n + 1];
        return min(1, n);
    }
}

min函数需要检查各种决策，而max函数是评判各种决策条件下的最坏情况。

简化的版本：

public class Solution {
    private int[][] amounts;
    private int min(int from, int to) {
        if (from >= to) return 0;
        if (amounts[from][to] > 0) return amounts[from][to];
        int min = Integer.MAX_VALUE;
        for(int i = from; i <= to; i++) {
            int cost = i + Math.max(min(from, i - 1), min(i + 1, to));
            min = Math.min(min, cost);
        }
        amounts[from][to] = min;
        return min;
    }
    public int getMoneyAmount(int n) {
        amounts = new int[n + 1][n + 1];
        return min(1, n);
    }
}