[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:

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

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)

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:

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++:

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++:

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++:

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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