LeetCode #464. Can I Win

最新推荐文章于 2021-07-13 23:19:44 发布

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本文探讨了100游戏的一种变种，其中两名玩家从1到15的数字池中不重复地抽取数字，目标是使总和达到或超过100。文章提供了判断先手玩家是否能确保胜利的算法实现，考虑了最大可选整数和期望总和的限制。

题目描述：

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.

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> visited;
        return can_win(maxChoosableInteger,desiredTotal,0,0,visited);
    }
    
    bool can_win(int max_int, int total, int used, int sum, unordered_map<int,bool>& visited)
    {
        if(visited.count(used)) return visited[used];
        for (int i=1;i<=max_int;i++) 
        {
            int cur=(1<<i);
            if((cur&used)==0) 
            {
                if (sum+i>=total||!can_win(max_int,total,used|cur,sum+i,visited)) 
                {
                    visited[used]=true;
                    return true;
                }
            }
        }
        visited[used]=false;
        return false;
    }
};