375. Guess Number Higher or Lower II

375. Guess Number Higher or Lower II

 

Question
Editorial Solution
 My Submissions

Total Accepted: 4164
Total Submissions: 13995
Difficulty: Medium

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.

这题求猜对数的最大花费的最小值。

通过模拟猜数过程得：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

class Solution {
public:
int cost(vector<vector<int> > &arr,int start,int end)
{
if (start>=end)
return 0;
if (arr[start][end]!=0)
return arr[start][end];
int res=INT_MAX;
for (int i=start;i<=end;++i)
{
int tmp=i+max(cost(arr,start,i-1),cost(arr,i+1,end));
if (tmp<res)
res=tmp;
}
arr[start][end]=res;
return res;
}
int getMoneyAmount(int n) {
vector<vector<int> > arr(n+1,vector<int>(n+1,0));
return cost(arr,1,n);
}
};

非递归代码：

int getMoneyAmount(int n) {
vector<vector<int> >cost(n+2,vector<int>(n+2,0));//下标0和n+1纯粹是为了计算方便而引入的，于是下面guess-1和guess+1无需判断是否越界
for(int diff=1;diff<=n;diff++){  //先循环距离，是因为刚开始的cost还没赋正确的值，你要先算完基本的。
for(int left=1;left+diff<=n;left++){
int right=left+diff;
int minCostInThisRange=INT_MAX;
for(int guess=left;guess<=right;guess++){
minCostInThisRange=min(minCostInThisRange,guess+max(cost[left][guess-1],cost[guess+1][right]));
}
cost[left][right]=minCostInThisRange;
}
}
return cost[1]
;
}