[LeetCode 293] Flip Game

最新推荐文章于 2022-08-30 23:45:03 发布

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探讨一个由+和-字符组成的字符串游戏，玩家通过翻转连续的两个'+', '++'为'--'来改变字符串状态。实现一个函数计算所有可能的一次有效操作后的字符串状态。

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 [].

solution: dfs

    @Test
    public void test() {
        String[] testcase = {"+","-","++","--","+++","++++","+--+","-+-+","+-+-"};
        for(int i=0;i<testcase.length;i++){
            System.out.println(getFlipResult(testcase[i]));
        }
    }
    public List<String> getFlipResult(String s1) {
        List<String> res = new ArrayList<String>();
        if(s1.length()<2) return res;
        dfs(s1, res, 0);
        return res;
    }
    public void dfs(String s1, List<String> res, int len) {
        for(int i=len;i<s1.length()-1;i++) {
            if(s1.charAt(i) == '+' && s1.charAt(i) == s1.charAt(i+1)) {
                String t1 = s1.substring(0, i) + "--" + s1.substring(i+2);
                res.add(t1);
            }else{
                dfs(s1, res, i+1);
            }
        }
    }