【Leetcode】Guess Number Higher or Lower II
2016-07-24 00:00
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题目链接:
题目:
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.Secondround:Youguess7,Itellyouthatit′shigher.Youpay7.
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.
思路:
关键在于把它转化为dp问题,O(n^3)的解法我总想不到。。/(ㄒoㄒ)/~~
设dp[i][j]为i~j要保证赢的最多代价。 假设猜k值,若不对,则问题转化为猜1~k-1,k+1~n两个子问题,且猜k的结果会告诉我们需要猜哪个子问题,所以猜k要保证赢 最多需要money=k+max{dp[i][k-1],dp[k+1,j]},则在i~j范围内存在某策略是最优的,则是猜测某个k需要的最少的money,为min{money} k属于[i,j]。
参考:https://discuss.leetcode.com/topic/51358/java-dp-solution
算法:
题目:
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.Secondround:Youguess7,Itellyouthatit′shigher.Youpay7.
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.
思路:
关键在于把它转化为dp问题,O(n^3)的解法我总想不到。。/(ㄒoㄒ)/~~
设dp[i][j]为i~j要保证赢的最多代价。 假设猜k值,若不对,则问题转化为猜1~k-1,k+1~n两个子问题,且猜k的结果会告诉我们需要猜哪个子问题,所以猜k要保证赢 最多需要money=k+max{dp[i][k-1],dp[k+1,j]},则在i~j范围内存在某策略是最优的,则是猜测某个k需要的最少的money,为min{money} k属于[i,j]。
参考:https://discuss.leetcode.com/topic/51358/java-dp-solution
算法:
public int getMoneyAmount(int n) { int[][] dp = new int[n + 1][n + 1]; for (int k = 1; k < n; k++) { for (int s = 1; s + k <= n; s++) { dp[s][s + k] = Integer.MAX_VALUE; for (int i = s; i < s + k; s++) { dp[s][s + k] = Math.min(dp[s][s + k], i + Math.max(dp[s][i - 1], dp[i + 1][s + k])); } } } return dp[1] ; }
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