LeetCode No.375 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.

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题目链接：https://leetcode.com/problems/guess-number-higher-or-lower-ii/

题目大意：求出保证能猜出结果的最小花费。

思路：使用动态规划。dp[i][j]表示从i到j的最小花费。则：

int ans = INT_MAX ;
for ( int i = begin ; i <= end ; i ++ )
{
    ans = min ( ans , i + max ( compute ( dp , begin , i - 1 ) , compute ( dp , i + 1 , end ) ) ) ;
}
dp[begin][end] = ans ;

参考代码：

class Solution {
public:
    int getMoneyAmount(int n) {
        if ( n <= 1 )
            return 0 ;
        int miner = 1 , maxer = n ;
        vector < vector <int> > dp ( n + 1 , vector <int> ( n + 1 , 0 ) ) ;
        return compute ( dp , 1 , n ) ;
    }
    int compute ( vector < vector <int> > & dp , int begin , int end )
    {
        if ( begin >= end )
            return 0 ;
        if ( dp[begin][end] )
            return dp[begin][end] ;
        int ans = INT_MAX ;
        for ( int i = begin ; i <= end ; i ++ )
        {
            ans = min ( ans , i + max ( compute ( dp , begin , i - 1 ) , compute ( dp , i + 1 , end ) ) ) ;
        }
        dp[begin][end] = ans ;
        return ans ;
    }
};