[LeetCode] 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.

自下而上的动态规划或者自上而下的记忆化dfs：

动态规划：

public class Solution2 {   public int getMoneyAmount(int n) { int[][] dp=new int[n+2][n+2];//这个小细节注意一下  因为Math.max(dp[i][k-1], dp[k+1][j])+k  所以两边都要留出默认值为0的空白边界 for(int len=0;len<n;len++){ for(int i=1;i<=n-len;i++){ int j=i+len; if(i==j) dp[i][j]=0; else{dp[i][j]=Integer.MAX_VALUE;for(int k=i;k<=j;k++){dp[i][j]=Math.min(dp[i][j],Math.max(dp[i][k-1], dp[k+1][j])+k);} } } } return dp[1][n]; }}

记忆化dfs：

public class Solution {    public int getMoneyAmount(int n) {         dp=new int[n+1][n+1];         for(int i=0;i<dp.length;i++){         Arrays.fill(dp[i],Integer.MAX_VALUE);         }         return dfs(1,n);    }        int[][] dp;    public int dfs(int l,int r){    if(l>=r) return 0;    if(dp[l][r]!=Integer.MAX_VALUE) return dp[l][r];    for(int i=l;i<=r;i++){    dp[l][r]=Math.min(dp[l][r], Math.max(i+dfs(l,i-1),i+dfs(i+1,r)));    }    return dp[l][r];    }}