leetcode 375. Guess Number Higher or Lower II 猜数游戏+动态规划DP+最大最小化问题
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.
在1-n个数里面，我们任意猜一个数(设为i)，保证获胜所花的钱应该为 i + max(w(1 ,i-1), w(i+1 ,n))，这里w(x,y))表示猜范围在(x,y)的数保证能赢应花的钱，则我们依次遍历 1-n作为猜的数，求出其中的最小值即为答案，即最小的最大值问题
建议和这道题leetcode 374. Guess Number Higher or Lower 猜数游戏一起学习
代码如下：
/*
* 这个事第二次遇到这样的DP动态规划问题
* 我们需要按照一定的顺序计算dp的值，由于计算dp[i][j]时我们需要计算dp[i][j-1],dp[i+1][j]，
* 所以我们按照j-i递增的顺序计算，即l区间长度len由短到长的顺序计算。
* */
class Solution
{
public int getMoneyAmount(int n)
{
int [][]dp=new int[n+1][n+1];
for(int len=1;lenfor(int i=0;i+len<=n;i++)
{
int j=i+len;
int one=Integer.MAX_VALUE;
for(int k=i;k<=j;k++)
{
int tmp=k+ Math.max(k-1>=i?dp[i][k - 1]:0,k+1<=j?dp[k + 1][j]:0);
one=Math.min(one, tmp);
}
dp[i][j]=one;
}
}
return dp[1][n];
//return cost(dp,1,n);
}
/*
* ，在1-n个数里面，我们任意猜一个数(设为i)，保证获胜所花的钱应该为 i + max(w(1 ,i-1), w(i+1 ,n))，
* 这里w(x,y))表示猜范围在(x,y)的数保证能赢应花的钱，则我们依次遍历 1-n作为猜的数，
* 求出其中的最小值即为答案，即最小的最大值问题
* */
public int cost(int[][] dp, int beg, int end)
{
if(beg>=end)
return 0;
else if(dp[beg][end]!=0)
return dp[beg][end];
else
{
int one=Integer.MAX_VALUE;
for(int i=beg;i<=end;i++)
{
int tmp=i+Math.max(cost(dp, beg, i-1), cost(dp, i+1, end));
one=Math.min(one, tmp);
}
dp[beg][end]=one;
return one;
}
}
} 下面是C++的做法，很不错，就是按照length去做DP动态规划
DFS的做法很不错，和这一道题leetcode 329. Longest Increasing Path in a Matrix 矩阵中寻找最长递增序列 + 一个典型的深度优先遍历DFS的做法 的做法很相似，建议一起学习
代码如下：
#include