[LeetCode] Flip Game

本文介绍了一个关于翻转游戏的算法实现,玩家通过翻转字符串中的连续++为--来获得胜利。文章提供了一种C++和Python的解决方案,通过遍历字符串并翻转符合条件的部分来生成所有可能的状态。代码示例清晰易懂,适合初学者学习。

Problem 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 compute all possible states of the string after one valid move.

For example, given s = "++++", after one move, it may become one of the following states:

[
  "--++",
  "+--+",
  "++--"
]

 

If there is no valid move, return an empty list [].


The idea is quite straightforward: just traverse s and each time when we see two consecutive+s, convert them to -s and add the resulting string to the final result moves. But remember to recover the string after that.

The C++ code is as follows.

class Solution {
public:
    vector<string> generatePossibleNextMoves(string s) {
        vector<string> moves;
        int n = s.length();
        for (int i = 0; i < n - 1; i++) {
            if (s[i] == '+' && s[i + 1] == '+') { 
                s[i] = s[i + 1] = '-';
                moves.push_back(s);
                s[i] = s[i + 1] = '+';
            }
        }
        return moves;
    }
};

Well I also try to write a Python solution since Python supports sequential comparisons, which is quite convenient. But Python does not support modifying a string and I can only use list and join to do the same thing.

class Solution(object):
    def generatePossibleNextMoves(self, s):
        """
        :type s: str
        :rtype: List[str]
        """
        moves, n, s = [], len(s), list(s)
        for i in xrange(n - 1):
            if s[i] == s[i + 1] == '+':
                s[i] = s[i + 1] = '-'
                moves += ''.join(s),
                s[i] = s[i + 1] = '+' 
        return moves

 

转载于:https://www.cnblogs.com/jcliBlogger/p/4882201.html

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