375. Guess Number Higher or Lower II - jiejackyzhang/leetcode-note GitHub Wiki

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.

解题思路为Dynamic Programming。

Definition of dp[i][j]: minimum number of money to guarantee win for subproblem [i, j].

Target: dp[1][n]

Corner case: dp[i][i] = 0 (because the only element must be correct)

Equation: we can choose k (i<=k<=j) as our guess, and pay price k. After our guess, the problem is divided into two subproblems. Notice we do not need to pay the money for both subproblems. We only need to pay the worst case (because the system will tell us which side we should go) to guarantee win. So dp[i][j] = min (i<=k<=j) { k + max(dp[i][k-1], dp[k+1][j]) }

解法与matrix chain multiplication类似。 这里有条件i<j，因此只需dp矩阵的右上对角，其余部分都为0，则满足base case，而且在dp[i]j时对结果也无影响。 这里设为dp[n+2][n+2]，是因为当j=n，k=j时，dp[k+1][j]不会溢出。

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