LeetCode375. Guess Number Higher or Lower II（动态规划）

Guess Number Higher or Lower II

Medium

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.

题意

从1到n中猜数，每次猜不中扣掉等于所猜数目的钱，猜中数字不扣钱。问使用最优策略下最坏情况至少准备多少钱，一定可以猜中。

思路

动态规划。

dp[i][j]

表示在

[i, j]

区间中进行猜中数字至少要准备的钱。边界情况

dp[i][i] = 0

（区间中只有一个数，那么结果就是这个数，一定能猜中，不用扣钱）

递推公式

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

其中

max

表示最坏情况，

min

表示最优策略

代码

class Solution {
    public int getMoneyAmount(int n) {
        int[][] dp = new int[n][n];
        int i = 0, j = 0, mid = 0;
        for (i=2; i<=n; ++i) {
            for (j=0; j<=n-i; ++j) {
                dp[j][i+j-1] = Integer.MAX_VALUE;
                dp[j][i+j-1] = Math.min(Math.min(dp[j][i+j-1], j+1 + dp[j+1][i+j-1]), i+j + dp[j][i+j-2]);
                for (mid = j+1; mid < i+j-1; ++mid) {
                    dp[j][i+j-1] = Math.min(dp[j][i+j-1], mid+1 + Math.max(dp[mid+1][i+j-1], dp[j][mid-1]));
                }
            }
        }
        return dp[0][n-1];
    }
}