# 293. Flip Game II

[Link](https://medium.com/leetcode-%E6%BC%94%E7%AE%97%E6%B3%95%E6%95%99%E5%AD%B8/013-leetcode-294-%E6%BC%94%E7%AE%97%E6%B3%95-flip-game-ii-%E7%BF%BB%E8%BD%89%E9%81%8A%E6%88%B2-99053b8d0ab5)

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.

**Solution**

用遞迴的方式解，只要存在下個變化被確定不能贏。則就代表目前的玩家一定能贏了。

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

```
