zoukankan      html  css  js  c++  java
  • [Leetcode] DP-- 464. Can I Win

    In the "100 game," two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins.

    What if we change the game so that players cannot re-use integers?

    For example, two players might take turns drawing from a common pool of numbers of 1..15 without replacement until they reach a total >= 100.

    Given an integer maxChoosableInteger and another integer desiredTotal, determine if the first player to move can force a win, assuming both players play optimally.

    You can always assume that maxChoosableInteger will not be larger than 20 and desiredTotal will not be larger than 300.

    Example

    Input:
    maxChoosableInteger = 10
    desiredTotal = 11
    
    Output:
    false
    
    Explanation:
    No matter which integer the first player choose, the first player will lose.
    The first player can choose an integer from 1 up to 10.
    If the first player choose 1, the second player can only choose integers from 2 up to 10.
    The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.
    Same with other integers chosen by the first player, the second player will always win.


    Solution: 
    1. the naive method is to get all the possible ways, which is factorial of maxChoosableInteger, maxChoosableInteger!.
     
    2.   use recursive and dp memoization to record the state
    state means which number has been visited by both. because there are maximum 20 for maxChoosableInteger,
    so we define a variable "visited" to record every bit as visited or not
    for different state, there is corresponding to win or not by player A, record as hash " winMap[state] " for memoization
    recursively call to judge win by drawing elements from 1 to maxChoosableInteger.
    when element could be drawed, at the same time left element > desiredTotal or the recursive call funtion,
     player A could win
     
       Then we consider the corner cases:
         if maxChoosableInteger  >desiredTotal:
              player A win
        if the sum of 1 tomaxChoosableInteger is bigger than
        player A loses
     
     1     winMap = {}
     2         def canIWinHelper(total, visited):
     3             if visited in winMap:
     4                 return winMap[visited]
     5             for i in range(1, maxChoosableInteger+1):
     6                 mask = (1 << i)
     7                 if (mask & visited) == 0 and (i >= total or not canIWinHelper(total - i, mask | visited)):
     8                     winMap[visited] = True
     9                     return True
    10             winMap[visited] = False
    11             return False
    12         
    13         if maxChoosableInteger > desiredTotal:
    14             return True
    15         if (1+maxChoosableInteger)*maxChoosableInteger/2 < desiredTotal:      #sum
    16             return False
    17         
    18         return canIWinHelper(desiredTotal, 0)
    19         
     
  • 相关阅读:
    C# WPF 窗体传递消息
    WPF ProgressBar 样式
    WPF的TextBox以及PasswordBox显示水印文字
    Win8.1 Hyper-V 共享本机IP上网
    ASP.NET MVC入门-Startup 类:注册服务和使用中间件
    ASP.NET MVC入门-Program类:程序的入口
    Unity3d AssetBundle 资源加载与管理
    C#考核知识点总结
    数据结构与算法之美-字符串匹配(上)
    LeetCode-探索链表-综合问题
  • 原文地址:https://www.cnblogs.com/anxin6699/p/7163176.html
Copyright © 2011-2022 走看看