[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 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.

For example, given s = "++++", return true. The starting player can guarantee a win by flipping the middle "++" to become "+--+".

Follow up:
Derive your algorithm's runtime complexity.

293. Flip Game 的拓展，这次求是否先玩者可以有一种策略一定赢游戏。

解法： backtracking

Java:

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public class Solution {     public boolean canWin(String s) {         for ( int i = 0; i < s.length() - 1; i ++ ){             if ( s.charAt ( i ) == '+' && s.charAt( i + 1 ) == '+' ){                 StringBuilder sb = new StringBuilder ( s );                 sb.setCharAt ( i , '-');                 sb.setCharAt ( i + 1 ,'-');                 if ( !canWin ( sb.toString() ) )                     return true;             }         }         return false;     } }

Java: backtracking

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public boolean canWin(String s) {     if(s==null||s.length()==0){         return false;     }       return canWinHelper(s.toCharArray()); }    public boolean canWinHelper(char[] arr){     for(int i=0; i<arr.length-1;i++){         if(arr[i]=='+'&&arr[i+1]=='+'){             arr[i]='-';             arr[i+1]='-';                boolean win = canWinHelper(arr);                arr[i]='+';             arr[i+1]='+';                //if there is a flip which makes the other player lose, the first play wins             if(!win){                 return true;             }         }     }        return false; }

 

Java:  DP, Time Complexity - O(2n)， Space Complexity - O(2n)

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public class Solution {     public boolean canWin(String s) {         char[] arr = s.toCharArray();         for(int i = 1; i < s.length(); i++) {             if(arr[i] == '+' && arr[i - 1] == '+') {                 arr[i] = '-';                 arr[i - 1] = '-';                 String next = String.valueOf(arr);                 if(!canWin(next)) {                     return true;                 }                 arr[i] = '+';                 arr[i - 1] = '+';             }         }                   return false;     } }

Python:

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class Solution(object):     def canWin(self, s):         """         :type s: str         :rtype: bool         """         for i in range(len(s) - 1): # 寻找所有的翻转可能             if s[i:i+2] == "++":                 current = s[0:i] + "--" + s[i+2:] # 把找到的++变成--                   if not self.canWin(current): # 看当前的字串是否存在边界，没有++了                     return True # 对手不能赢，那就是当前翻转的赢了         return False # loop中没有返回，不能赢，当前翻转的输了

C++:

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class Solution { public:     bool canWin(string s) {         for (int i = 1; i < s.size(); ++i) {             if (s[i] == '+' && s[i - 1] == '+' && !canWin(s.substr(0, i - 1) + "--" + s.substr(i + 1))) {                 return true;             }         }         return false;     } };

C++:

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class Solution { public:     bool canWin(string s) {    //朴素回溯，715MS         int len=s.size();         if(len<=1) return false;         for(int i=0;i<len-1;i++) {             string tmp=s;             if(s[i]=='+'&&s[i+1]=='+') {                 tmp[i]='-';tmp[i+1]='-';                 bool f=canWin(tmp);                 if(!f) return true;             }         }         return false;     } };

C++:

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class Solution { public:     bool canWin(string s) {    //记录中间结果，39MS         int len=s.size();         if(len<=1) return false;         if(Memory_Map.find(s)!=Memory_Map.end()) {             return Memory_Map[s];         }         for(int i=0;i<len-1;i++) {             string tmp=s;             if(s[i]=='+'&&s[i+1]=='+') {                 tmp[i]='-';tmp[i+1]='-';                 bool f=canWin(tmp);                 if(!f) {                     Memory_Map[s]=true;                     return true;                 }             }         }         Memory_Map[s]=false;         return false;     } private:     unordered_map<string,bool> Memory_Map; };

　　

 

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