LeetCode-Flip Game II

Posted 2020-12-26 IncredibleThings

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.

若是有一段"++", 剩下的段和"--"组合 can not win, 那么返回true.

从头到尾试遍了没找到这么一段"++", 返回false.

Time Complexity: exponential.

T(n) = (n-2) * T(n-2) = (n-2) * (n-4) * T(n-4) = O(n!!)

public class Solution {
    public boolean canWin(String s) {
        for(int i = 1; i<s.length(); i++){
            if(s.charAt(i) == ‘+‘ && s.charAt(i-1) == ‘+‘ && !canWin(s.substring(0, i-1) + "--" + s.substring(i+1))){
                return true;
            }
        }
        return false;
    }
}