You are playing the following Flip Game with your friend: Given a string that contains only these two characters: + and -, you and your friend take turns to flip two consecutive "++" into "--". The game ends when a person can no longer make a move and therefore the other person will be the winner.

Write a function to determine if the starting player can guarantee a win.

Example:

Input: s = "++++"
Output: true
Explanation: The starting player can guarantee a win by flipping the middle "++" to become "+--+".

Follow up:
Derive your algorithm's runtime complexity.

这道题是之前那道 Flip Game 的拓展,让我们判断先手的玩家是否能赢,可以穷举所有的情况,用回溯法来解题,思路跟上面那题类似,也是从第二个字母开始遍历整个字符串,如果当前字母和之前那个字母都是+,那么递归调用将这两个位置变为--的字符串,如果返回 false,说明当前玩家可以赢,结束循环返回 false。这里同时贴上热心网友 iffalse 的解释,这道题不是问 “1p是否会怎么选都会赢”,而是 “如果1p每次都选特别的两个+,最终他会不会赢”。所以 canWin 这个函数的意思是 “在当前这种状态下,至少有一种选法,能够让他赢”。而 (!canWin) 的意思就变成了 “在当前这种状态下,无论怎么选,都不能赢”。所以 1p 要看的是,是否存在这样一种情况,无论 2p 怎么选,都不会赢。所以只要有一个 (!canWin),1p 就可以确定他会赢。这道题从博弈论的角度会更好理解。每个 player 都想让自己赢,所以每轮他们不会随机选+。每一轮的 player 会选能够让对手输的+。如果无论如何都选不到让对手输的+,那么只能是当前的 player 输了,参见代码如下:

解法一:

class Solution {
public:
bool canWin(string s) {
for (int i = ; i < s.size(); ++i) {
if (s[i] == '+' && s[i - ] == '+' && !canWin(s.substr(, i - ) + "--" + s.substr(i + ))) {
return true;
}
}
return false;
}
};

第二种解法和第一种解法一样,只是用 find 函数来查找 ++ 的位置,然后把位置赋值给i,然后还是递归调用 canWin 函数,参见代码如下:

解法二:

class Solution {
public:
bool canWin(string s) {
for (int i = -; (i = s.find("++", i + )) >= ;) {
if (!canWin(s.substr(, i) + "--" + s.substr(i + ))) {
return true;
}
}
return false;
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/294

类似题目:

Nim Game

Flip Game

Guess Number Higher or Lower II

Can I Win

参考资料:

https://leetcode.com/problems/flip-game-ii/

https://leetcode.com/problems/flip-game-ii/discuss/74033/4-line-Java-Solution

https://leetcode.com/problems/flip-game-ii/discuss/74010/Short-Java-and-Ruby

https://leetcode.com/problems/flip-game-ii/discuss/73962/Share-my-Java-backtracking-solution

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] Flip Game II 翻转游戏之二的更多相关文章

  1. [LeetCode] Jump Game II 跳跃游戏之二

    Given an array of non-negative integers, you are initially positioned at the first index of the arra ...

  2. [LeetCode] Flip Game 翻转游戏之二

    You are playing the following Flip Game with your friend: Given a string that contains only these tw ...

  3. [LeetCode] 45. Jump Game II 跳跃游戏之二

    Given an array of non-negative integers, you are initially positioned at the first index of the arra ...

  4. [LeetCode] Reverse String II 翻转字符串之二

    Given a string and an integer k, you need to reverse the first k characters for every 2k characters ...

  5. [Swift]LeetCode294. 翻转游戏之 II $ Flip Game II

    You are playing the following Flip Game with your friend: Given a string that contains only these tw ...

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

    You are playing the following Flip Game with your friend: Given a string that contains only these tw ...

  7. [LeetCode] Bulb Switcher II 灯泡开关之二

    There is a room with n lights which are turned on initially and 4 buttons on the wall. After perform ...

  8. [LeetCode] Word Pattern II 词语模式之二

    Given a pattern and a string str, find if str follows the same pattern. Here follow means a full mat ...

  9. [LeetCode] Paint House II 粉刷房子之二

    There are a row of n houses, each house can be painted with one of the k colors. The cost of paintin ...

随机推荐

  1. poj3669 Meteor Shower(预处理+bfs)

    https://vjudge.net/problem/POJ-3669 先给地图a[][]预处理每个位置被砸的最小时间.然后再bfs. 纯bfs,还被cin卡了下时间.. #include<io ...

  2. 使用C#版Tesseract库

    上一篇介绍了Tesseract库的使用(OCR库Tesseract初探),文末提到了Tesseract是用c/c++开发的,也有C#的开源版本,本篇介绍一下如何使用C#版的Tesseract. C#版 ...

  3. thymeleaf学习笔记

    1.${@dict.hello().fatherName} 显示对象的属性2.${@dict.hello()[0].fatherName} 显示列表对象的属性3.<div th:object=& ...

  4. Jquery如何获取某个元素前(后)的文本内容?

    <span> text here... <a id="target_element">百万创想</a></span> 如何获得a标签 ...

  5. npm run dev

     npm run dev     npm run build

  6. phpBB3导入用户的Python脚本

    关联的数据表 在phpBB3中导入用户时, 需要处理的有两张表, 一个是 users, 一个是 user_group. 如果是新安装的论坛, 在每次导入之前, 用以下语句初始化: DELETE FRO ...

  7. [转]The Production Environment at Google

    A brief tour of some of the important components of a Google Datacenter.   A photo of the interior o ...

  8. 10分钟上手图数据库Neo4j

    随着互联网不断的发展,传统的关系型数据库如oracle,mysql已经难以支撑现下大数据量,高并发的场景了.于是,NoSQL横空出世,有像cassandra这样的column-based,像Mongo ...

  9. C#7.2——编写安全高效的C#代码 c# 中模拟一个模式匹配及匹配值抽取 走进 LINQ 的世界 移除Excel工作表密码保护小工具含C#源代码 腾讯QQ会员中心g_tk32算法【C#版】

    C#7.2——编写安全高效的C#代码 2018-11-07 18:59 by 沉睡的木木夕, 123 阅读, 0 评论, 收藏, 编辑 原文地址:https://docs.microsoft.com/ ...

  10. Effective Java 第三版——67. 明智谨慎地进行优化

    Tips 书中的源代码地址:https://github.com/jbloch/effective-java-3e-source-code 注意,书中的有些代码里方法是基于Java 9 API中的,所 ...