标签:make class you min into bec move cte character
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; } }
标签:make class you min into bec move cte character
原文地址:https://www.cnblogs.com/incrediblechangshuo/p/9388359.html