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Leetcode 294. Flip GameII

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Question:

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.

Explanation:

本来想找规律，后来想想太天真了。用backtracking解，罗列所有的情况，如果下一局对手会输，就表示可以。现在我的算法是o(N!!) discuss中有一个牛逼的game theory博弈论算法，看不懂。有机会可以研究一下，会将time complexity降低到O(N^2)

Code:

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public class Solution {
    public boolean canWin(String s) {
        if (s == null || s.length() <2) return false;
        char[] charString = s.toCharArray();
        for (int i=0; i<s.length()-1; i++) {
            if (charString[i] == '+' && charString[i+1] == '+') {
                charString[i] = '-';
                charString[i+1] = '-';
                if (!canWin(String.valueOf(charString))) {
                    return true;
                }
                charString[i] = '+'; charString[i+1] = '+';
            }
        }
        return false;
    }
}