leetcode 374&375. Guess Number Higher or Lower

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 is higher or lower.

You call a pre-defined API guess(int num) which returns 3 possible results (-1, 1, or 0):

-1 : My number is lower
 1 : My number is higher
 0 : Congrats! You got it!
Example:
n = 10, I pick 6.

Return 6.

374比较简单，就是一个二分搜索

public class Solution extends GuessGame {
    public int guessNumber(int n) {
        int low = 1;
        int high = n;
        int mid = 0;
        while(low <= high){
            mid = low + (high - low)/2;
            int res = guess(mid);
            if(res==0){
                return mid;
            }else if(res>0){
                low = mid+1;
            }else{
                high = mid-1;
            }
        }
        return low;
    }
}

375二分搜索不一定是最好的结果，所以不能采用二分搜索来求。
采用动态规划
dp[i][j]表示从i~j中的最少花费
对于 x 属于i~j，如果是x，那么花费为x + max(dp[i][x-1], dp[x+1][j])

在最小化这个花费即为dp[i][j]

public class Solution {
    public int getMoneyAmount(int n) {
        if(n < 2) return 0;
        int[][] dp = new int[n+1][n+1];
        return helper(dp, 1, n);

    }

    public int helper(int[][] dp, int l, int h){
        if(l >= h) return 0;
        if(dp[l][h]!=0) return dp[l][h];

        int res = Integer.MAX_VALUE;
        for(int i=l; i <=h; i++){
            int t = i + Math.max(helper(dp, l, i-1), helper(dp, i+1, h));
            res = Math.min(res, t);
        }

        dp[l][h] = res;
        return res;
    }
}