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

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10692494.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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.

---

您正在和您的朋友玩以下翻转游戏：给定一个仅包含这两个字符的字符串：+和-，您和您的朋友轮流将两个连续的“++”翻转为“-”。游戏结束时，一个人不能再做一个动作，因此另一个人将是赢家。

写一个函数来确定首发球员是否能保证获胜。

例子：

输入：s = "++++"

输出：true

说明：先手玩家可以通过将中间“++”翻转成“+--+”来保证获胜。

跟进：

推导算法的运行时复杂性。

---

Solution:

class Solution {
    func canWin(_ s:String) -> Bool {
        var arrS:[Character] = Array(s)
        for i in 1..<s.count
        {
            if arrS[i] == "+" && arrS[i - 1] == "+" && !canWin(s.subString(0, i - 1) + "--" + s.subString(i + 1))
            {
                return true
            }
        }
        return false          
    }
}

extension String {
    // 截取字符串：从index到结束处
    // - Parameter index: 开始索引
    // - Returns: 子字符串
    func subString(_ index: Int) -> String {
        let theIndex = self.index(self.endIndex, offsetBy: index - self.count)
        return String(self[theIndex..<endIndex])
    }
    
    // 截取字符串：指定索引和字符数
    // - begin: 开始截取处索引
    // - count: 截取的字符数量
    func subString(_ begin:Int,_ count:Int) -> String {
        let start = self.index(self.startIndex, offsetBy: max(0, begin))
        let end = self.index(self.startIndex, offsetBy:  min(self.count, begin + count))
        return String(self[start..<end]) 
    }
}

---

点击：Playground测试

var sol = Solution()
print(sol.canWin("++++"))
//Print true