zoukankan      html  css  js  c++  java
  • [LeetCode] 464. Can I Win 我能赢吗

    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.

    给了一组数字,两个人玩游戏,每人每次可选一个数字,看数字总和谁先到给定值,判断第一玩的人能否保证能赢。

    Java:

    public class Solution {
        Map<Integer, Boolean> map;
        boolean[] used;
        public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
            int sum = (1+maxChoosableInteger)*maxChoosableInteger/2;
            if(sum < desiredTotal) return false;
            if(desiredTotal <= 0) return true;
            
            map = new HashMap();
            used = new boolean[maxChoosableInteger+1];
            return helper(desiredTotal);
        }
        
        public boolean helper(int desiredTotal){
            if(desiredTotal <= 0) return false;
            int key = format(used);
            if(!map.containsKey(key)){
        // try every unchosen number as next step
                for(int i=1; i<used.length; i++){
                    if(!used[i]){
                        used[i] = true;
         // check whether this lead to a win (i.e. the other player lose)
                        if(!helper(desiredTotal-i)){
                            map.put(key, true);
                            used[i] = false;
                            return true;
                        }
                        used[i] = false;
                    }
                }
                map.put(key, false);
            }
            return map.get(key);
        }
       
    // transfer boolean[] to an Integer 
        public int format(boolean[] used){
            int num = 0;
            for(boolean b: used){
                num <<= 1;
                if(b) num |= 1;
            }
            return num;
        }
    }
    

    Python:  

    def canIWin(self, maxChoosableInteger, desiredTotal):
            """
            :type maxChoosableInteger: int
            :type desiredTotal: int
            :rtype: bool
            """
            if (1 + maxChoosableInteger) * maxChoosableInteger/2 < desiredTotal:
                return False
            self.memo = {}
            return self.helper(range(1, maxChoosableInteger + 1), desiredTotal)
    
            
        def helper(self, nums, desiredTotal):
            
            hash = str(nums)
            if hash in self.memo:
                return self.memo[hash]
            
            if nums[-1] >= desiredTotal:
                return True
                
            for i in range(len(nums)):
                if not self.helper(nums[:i] + nums[i+1:], desiredTotal - nums[i]):
                    self.memo[hash]= True
                    return True
            self.memo[hash] = False
            return False
    

    Python:  

    # Memoization solution.
    class Solution(object):
        def canIWin(self, maxChoosableInteger, desiredTotal):
            """
            :type maxChoosableInteger: int
            :type desiredTotal: int
            :rtype: bool
            """
            def canIWinHelper(maxChoosableInteger, desiredTotal, visited, lookup):
                if visited in lookup:
                    return lookup[visited]
    
                mask = 1
                for i in xrange(maxChoosableInteger):
                    if visited & mask == 0:
                        if i + 1 >= desiredTotal or 
                           not canIWinHelper(maxChoosableInteger, desiredTotal - (i + 1), visited | mask, lookup):
                            lookup[visited] = True
                            return True
                    mask <<= 1
                lookup[visited] = False
                return False
    
            if (1 + maxChoosableInteger) * (maxChoosableInteger / 2) < desiredTotal:
                return False
    
            return canIWinHelper(maxChoosableInteger, desiredTotal, 0, {})  

    C++:

    class Solution {
    public:
        bool canIWin(int maxChoosableInteger, int desiredTotal) {
            if (maxChoosableInteger >= desiredTotal) return true;
            if (maxChoosableInteger * (maxChoosableInteger + 1) / 2 < desiredTotal) return false;
            unordered_map<int, bool> m;
            return canWin(maxChoosableInteger, desiredTotal, 0, m);
        }
        bool canWin(int length, int total, int used, unordered_map<int, bool>& m) {
            if (m.count(used)) return m[used];
            for (int i = 0; i < length; ++i) {
                int cur = (1 << i);
                if ((cur & used) == 0) {
                    if (total <= i + 1 || !canWin(length, total - (i + 1), cur | used, m)) {
                        m[used] = true;
                        return true;
                    }
                }
            }
            m[used] = false;
            return false;
        }
    };
    

      

    类似题目:

    486. Predict the Winner

    375. Guess Number Higher or Lower II

    [LeetCode] 294. Flip Game II 翻转游戏 II

    877. Stone Game

    All LeetCode Questions List 题目汇总

  • 相关阅读:
    POJ 1426 Find The Multiple (DFS / BFS)
    C++的class的样例
    HDU 5063 Operation the Sequence(暴力 数学)
    VmWare 与 宿主主机通信 STEP BY STEP (适用于刚開始学习的人)
    Pig源代码分析: 简析运行计划的生成
    Extjs, 使用GridPanel出现 Layout run failed
    Codeforces #258 Div.2 E Devu and Flowers
    pat(A) 2-06. 数列求和(模拟摆竖式相加)
    javaWeb Cache技术――OSCache(转-全)
    108句个性搞笑签名 (转)
  • 原文地址:https://www.cnblogs.com/lightwindy/p/9736348.html
Copyright © 2011-2022 走看看