LeetCode No.375 Guess Number Higher or Lower II
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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 ; }};
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