[LeetCode] Flip Game 翻转游戏

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 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 [].

这道题让我们把相邻的两个++变成--，真不是一道难题，我们就从第二个字母开始遍历，每次判断当前字母是否为+，和之前那个字母是否为+，如果都为加，则将翻转后的字符串存入结果中即可，参见代码如下：

class Solution {
public:
    vector<string> generatePossibleNextMoves(string s) {
        vector<string> res;
        for (int i = 1; i < s.size(); ++i) {
            if (s[i] == '+' && s[i - 1] == '+') {
                res.push_back(s.substr(0, i - 1) + "--" + s.substr(i + 1));
            }
        }
        return res;
    }
};

本文转自博客园Grandyang的博客，原文链接：翻转游戏[LeetCode] Flip Game ，如需转载请自行联系原博主。