★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號:山青詠芝(shanqingyongzhi)
➤博客園地址:山青詠芝(https://www.cnblogs.com/strengthen/)
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:http://www.javashuo.com/article/p-buzevrts-kv.html
➤若是連接不是山青詠芝的博客園地址,則多是爬取做者的文章。
➤原文已修改更新!強烈建議點擊原文地址閱讀!支持做者!支持原創!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html
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.git
Write a function to determine if the starting player can guarantee a win.github
Example:算法
Input: Output: true Explanation: The starting player can guarantee a win by flipping the middle to become . s = "++++""++""+--+"
Follow up:
Derive your algorithm's runtime complexity.微信
您正在和您的朋友玩如下翻轉游戲:給定一個僅包含這兩個字符的字符串:+和-,您和您的朋友輪流將兩個連續的「++」翻轉爲「-」。遊戲結束時,一我的不能再作一個動做,所以另外一我的將是贏家。函數
寫一個函數來肯定首發球員是否能保證獲勝。測試
例子:spa
輸入:s = "++++"
code
輸出:truehtm
說明:先手玩家能夠經過將中間「++」翻轉成「+--+」來保證獲勝。
跟進:
推導算法的運行時複雜性。
Solution:
1 class Solution { 2 func canWin(_ s:String) -> Bool { 3 var arrS:[Character] = Array(s) 4 for i in 1..<s.count 5 { 6 if arrS[i] == "+" && arrS[i - 1] == "+" && !canWin(s.subString(0, i - 1) + "--" + s.subString(i + 1)) 7 { 8 return true 9 } 10 } 11 return false 12 } 13 } 14 15 extension String { 16 // 截取字符串:從index到結束處 17 // - Parameter index: 開始索引 18 // - Returns: 子字符串 19 func subString(_ index: Int) -> String { 20 let theIndex = self.index(self.endIndex, offsetBy: index - self.count) 21 return String(self[theIndex..<endIndex]) 22 } 23 24 // 截取字符串:指定索引和字符數 25 // - begin: 開始截取處索引 26 // - count: 截取的字符數量 27 func subString(_ begin:Int,_ count:Int) -> String { 28 let start = self.index(self.startIndex, offsetBy: max(0, begin)) 29 let end = self.index(self.startIndex, offsetBy: min(self.count, begin + count)) 30 return String(self[start..<end]) 31 } 32 }
點擊:Playground測試
1 var sol = Solution() 2 print(sol.canWin("++++")) 3 //Print true