小旭讲解 LeetCode 464. Can I Win
小旭讲解 LeetCode 464. Can I Win

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原题

中文

在 “100 game” 这个游戏中,两名玩家轮流选择从 1 到 10 的任意整数,累计整数和,先使得累计整数和达到 100 的玩家,即为胜者。

如果我们将游戏规则改为 “玩家不能重复使用整数” 呢?

例如,两个玩家可以轮流从公共整数池中抽取从 1 到 15 的整数(不放回),直到累计整数和 >= 100。

给定一个整数 maxChoosableInteger (整数池中可选择的最大数)和另一个整数 desiredTotal(累计和),判断先出手的玩家是否能稳赢(假设两位玩家游戏时都表现最佳)?

你可以假设 maxChoosableInteger 不会大于 20, desiredTotal 不会大于 300。

英文

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.

思路

小旭讲解 LeetCode 464. Can I Win
小旭讲解 LeetCode 464. Can I Win
小旭讲解 LeetCode 464. Can I Win
小旭讲解 LeetCode 464. Can I Win

代码

class Solution {
private:
    bool helper(vector<int8_t>& vis, int& mvalue, int num, int total) {
        if (~vis[num]) return vis[num];
       
        vis[num] = 0;
        for (int i = 1; i <= mvalue; ++i) {
            if (num & (1 << (i - 1))) continue;
            
            if (i >= total) {
                vis[num] = 1;
                break;
            }
            
            bool next = helper(vis, mvalue, num | (1 << (i - 1)), total - i);
            if (next == false) {
                vis[num] = 1;
                break;
            }
        }
        
        return vis[num];
    };
public:
    bool canIWin(int mvalue, int total) {
        if ((mvalue * (mvalue + 1) / 2) < total) return false;
        vector<int8_t> vis((1 << mvalue) - 1, -1);
        int num = 0;
        return helper(vis, mvalue, num, total);
    }
};

复杂度分析

时间复杂度:O((1 << M) - 1) M for maxChoosableInteger  不严谨
空间复杂度:O((1 << M) - 1)

参考

原题链接: https://leetcode.com/problems/can-i-win/

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本文地址:https://qoogle.top/leetcode-464-can-i-win/
最后修改日期:2020年6月18日

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