[LeetCode] Flip Game 翻转游戏之二

Posted 2020-06-17

 

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 twoconsecutive "++" 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.

 

这道题是之前那道Flip Game的拓展，让我们判断先手的玩家是否能赢，那么我们可以穷举所有的情况，用回溯法来解题，我们的思路跟上面那题类似，也是从第二个字母开始遍历整个字符串，如果当前字母和之前那个字母都是+，那么我们递归调用将这两个位置变为--的字符串，如果返回false，说明当前玩家可以赢，结束循环返回false，参见代码如下：

 

解法一：

class Solution {
public:
    bool canWin(string s) {
        for (int i = 1; i < s.size(); ++i) {
            if (s[i] == ‘+‘ && s[i - 1] == ‘+‘ && !canWin(s.substr(0, i - 1) + "--" + s.substr(i + 1))) {
                return true;
            }
        }
        return false;
    }
};

 

第二种解法和第一种解法一样，只是用find函数来查找++的位置，然后把位置赋值给i，然后还是递归调用canWin函数，参见代码如下：

 

解法二：

class Solution {
public:
    bool canWin(string s) {
        for (int i = -1; (i = s.find("++", i + 1)) >= 0;) {
            if (!canWin(s.substr(0, i) + "--" + s.substr(i + 2))) {
                return true;
            }
        }
        return false;
    }
};

 

类似题目：

Flip Game

 

参考资料：

https://leetcode.com/discuss/64350/short-java-%26-ruby

https://leetcode.com/discuss/64330/4-line-java-solution

 

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