Leetcode-464. Can I Win

最新推荐文章于 2023-02-25 23:03:36 发布

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最新推荐文章于 2023-02-25 23:03:36 发布

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分类专栏：算法文章标签： java leetcode 算法

本文链接：https://blog.youkuaiyun.com/mcf171/article/details/53992864

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这篇博客探讨了LeetCode中的464题，即在一个不允许重复使用数字的100游戏中，首位玩家是否能通过优化策略确保胜利。博主分享了他们的解题思路，并欢迎读者交流讨论。题目中指出，两位玩家从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。在给定的限制条件下，需要判断首位玩家是否有必胜策略。

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发优快云，mcf171专栏。

博客链接：mcf171的博客

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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.

最近这几个题确实挺蛋疼的。这个问题一直在想是否存在最优解，但是后来发现还是不如暴力解，做题目最伤的就是好不容易写出来，然后TTL。 Your runtime beats 52.71% of java submissions.

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;
    }
}