375. Guess Number Higher or Lower II
2017-05-03 11:00
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LeetCode
题目地址:https://leetcode.com/problems/guess-number-higher-or-lower-ii/#/description问题描述&解题思路:告诉你一个数n,要你从1~n中猜出对方选的数,每次的代价是你猜的那个数,对方会告诉你大了还是小了或者猜对了,猜对的那次不需要付代价。一开始想着就二分呗,每次选大的那边继续,但发现好像不是这么简单比如n=4的时候,最小代价应该是1+3,而n=6的时候最小代价应该是3+5,并不是简单的二分。于是就用动态规划来求解,用v[i][j]表示猜出i到j需要付出的最小代价,那么需要准备的事情如下:
初始状态:i >= j的时候,v[i][j]=0
状态转移:i < j的时候,v[i][j]=min(k+max(v[i][k-1],v[k+1][j])),其中k从i到j。这里稍微解释一下,k表示本次猜的数可选范围是i到j,而v[i][k-1]和v[k+1][j]是两个子问题,所以只需要把能猜的值都猜一遍,并且找到其中最小的代价作为v[i][j],上面的min就是这个意思。而max则是考虑最坏情况,k选取了分支之后,max表示最坏情况,即代价最大的分支需要付出的代价。
代码如下:
class Solution { public: int getMoneyAmount(int n) { vector<vector<int>> v(n,vector<int>(n,-1)); int max = DP(v,0,n-1); for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { cout << i+1 << "," << j+1 << " : " << v[i][j] << endl; } } return max; } int DP(vector<vector<int>> &v, int begin, int end) { if (begin >= end) return 0; if (v[begin][end] != -1) return v[begin][end]; int min = INT_MAX; int minIndex; for (int i = begin; i <= end; i++ ) { int tmp = (i+1) + max(DP(v,begin,i-1),DP(v,i+1,end)); if (tmp < min) { min = tmp; minIndex = i+1; } } v[begin][end] = min; return min; } };
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