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# Flip Game II

<https://leetcode.com/problems/flip-game-ii/description/>

> 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.

## Thoughts

可以用博弈论理论得出O(N^2)的解, 但面试时并不会考察. 用DFS + Memorization就可以了.

用递归模拟两个人的玩游戏的步骤, 并把结果存下来. 如果对手步骤返回false, 说明对手无地可下, 当前玩家的步骤返回true.

## Code

```
class Solution {
    private boolean canWin(char[] str, Map<String, Boolean> cache) {
        String ss = new String(str);
        if (cache.containsKey(ss)) {
            return cache.get(ss);
        }
        for (int i = 0; i < str.length - 1; i++) {
            if (str[i] == '+' && str[i + 1] == '+') {
                str[i] = '-';
                str[i + 1] = '-';
                boolean res = !canWin(str, cache);
                str[i] = '+';
                str[i + 1] = '+';
                if (res) {
                    cache.put(ss, true);
                    return true;
                }
            }
        }
        cache.put(ss, false);
        return false;
    }

    public boolean canWin(String s) {
        return canWin(s.toCharArray(), new HashMap<>());
    }
}
```

## Analysis

Errors: 1. 不能!canWin(str)直接返回true. 要把该递归调用修改过的str全部改回来.

如果不加memory, T(N) = (N - 2) *T(N - 4) = (N - 2)* (N - 4) *T(N - 6) = .... 由Double factorial可知时间复杂度为O(N!!), 即2 ^ N* (N - 1). 加了memory后未知.