leetcode 375. Guess Number Higher or Lower II

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375. Guess Number Higher or Lower II

 

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• Total Accepted: 1546
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• 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

题解转自：

http://blog.csdn.net/adfsss/article/details/51951658

题目大意还是猜数游戏，不过这次的重点换了，题目要求你猜错了就得付与你所猜的数目一致的钱，最后应求出保证你能赢的钱数(即求出保证你获胜的最少的钱数)，题目的提示是动态规划，自己吃的不太透，想着之前做过的动态规划都是从初始条件出发，再推出后面的值，于是就噗嗤噗嗤地写了前面几种情况在找规律，算了下，发现max(后四位的所应判断的钱数,倒数第四位的钱数+前面的钱数)好像挺有可能的，但提交的时候发现n大于19时就统统过不了了，于是果断否决了自己的想法，因为找规律这个想法完全说服不了自己为什么这样求出的是最小值，在这里也给自己敲了一个警钟，不要试图用侥幸的心理来蒙骗自己，网上了搜一番后，发现没有讲的比较清楚的博客，统统都是贴代码(っ*´Д`)っ(理解不了了喂)，后面在leetcode的discuss里面有两句英文让自己一下子就懂了.

具体是这样的，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

转：

C++ vector多维数组初始化及清零

http://blog.csdn.net/xiaxiazls/article/details/50018225

定义并初始化二维数组

vector<vector <int> > ivec(m ,vector<int>(n,0)); //m*n的二维vector，所有元素初始化为0

class Solution {
public:
    int getMoneyAmount(int n) {
        //vector<vector <int> > ivec(m ,vector<int>(n,0)); //m*n的二维vector，所有元素初始化为0
        vector<vector <int> > dp(n + 1 ,vector<int>(n + 1, INT_MAX)); //n*n的二维vector，所有元素初始化为INT_MAX
        dfs(dp,1,n);
        return dp[1][n];
    }
    int dfs(vector<vector <int> > &dp,int l,int r){
        if(dp[l][r] != INT_MAX){
            return dp[l][r];
        }
        if(l == r){
            return dp[l][r] = 0;
        }
        if(l + 1 == r){
            return dp[l][r] = l;
        }
        int i;
        for(i = l + 1;i <= r - 1;i++){
            dp[l][r] = min(dp[l][r],i + max(dfs(dp,l,i - 1),dfs(dp,i + 1,r) ) );
        }
        return dp[l][r];
    }
};