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 [].

一刷：
由于是输出一次move后的所有状态。
思路很简单，从0找到“++”所在的index i, 然后替换（在res中直接添加）为“--”，然后从i之后继续寻找“++ ”

public class Solution {
    public List<String> generatePossibleNextMoves(String s) {
        List<String> list = new ArrayList<>();
        for(int i=-1; (i = s.indexOf("++", i+1))>=0;)
            list.add(s.substring(0, i) + "--" + s.substring(i+2));
        return list;
    }
}

二刷
遍历去寻找连续的“+”

public class Solution {
    public List<String> generatePossibleNextMoves(String s) {
        List<String> list = new ArrayList<>();
        for(int i=0; i<s.length()-1; i++){
            if(s.charAt(i) == '+' && s.charAt(i+1) == '+'){
                StringBuilder sb = new StringBuilder();
                sb.append(s.substring(0, i)).append("--").append(s.substring(i+2));
                list.add(sb.toString());
            }
        }
        return list;
    }
}

三刷

class Solution {
    public List<String> generatePossibleNextMoves(String s) {
        List<String> res = new ArrayList<>();
        for(int i=0; i<s.length()-1; i++){
            char[] chs = s.toCharArray();
            if(chs[i] == '+' && chs[i+1] == '+'){
                chs[i] = '-';
                chs[i+1] = '-';
                res.add(new String(chs));
            }
        }
        return res;
    }
}