294. Flip Game 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.

本题采用回溯法，什么是回溯法呢？回溯法是可以系统地搜索原问题的所有解或者任意解，采用DFS方法进行的搜索，有剪枝函数和限界函数。本问题难点在于转换。因为有两个人进行比赛，第一个人操作完，第二个人进行判断，如果是false的话，那么第一个人就会取胜。本题还有一个难点就是添加一个Hashset来进行memorization的操作，用于存储出现重复的并且会取胜的String，代码如下：

public class Solution {
    public boolean canWin(String s) {
        if(s==null||s.length()<2) return false;
        Set<String> set = new HashSet<String>();
        return canWin(s,set);
    }
    public boolean canWin(String s,Set<String> set){
        if(set.contains(s)) return true;
        for(int i=0;i<s.length()-1;i++){
            if(s.startsWith("++",i)){
                String t = s.substring(0,i)+"--"+s.substring(i+2);
                if(!canWin(t,set)){
                    set.add(s);
                    return true;
                }
            }
        }
        return false;
    }
}

 答案做了进一步的改进，代码如下：

public class Solution {
    public boolean canWin(String s) {
        if(s==null||s.length()<2) return false;
        Map<String,Boolean> map = new HashMap<String,Boolean>();
        return canWin(s,map);
    }
    public boolean canWin(String s,Map<String,Boolean> map){
        if(map.containsKey(s)) return map.get(s);
        for(int i=0;i<s.length()-1;i++){
            if(s.startsWith("++",i)){
                String t = s.substring(0,i)+"--"+s.substring(i+2);
                if(!canWin(t,map)){
                    map.put(s,true);
                    return true;
                }
            }
        }
        map.put(s,false);
        return false;
    }
}