LeetCode 777. Swap Adjacent in LR String

In a string composed of ‘L’, ‘R’, and ‘X’ characters, like “RXXLRXRXL”, a move consists of either replacing one occurrence of “XL” with “LX”, or replacing one occurrence of “RX” with “XR”. Given the starting string start and the ending string end, return True if and only if there exists a sequence of moves to transform one string to the other

Example:

Input: start = "RXXLRXRXL", end = "XRLXXRRLX"
Output: True

Explanation:
We can transform start to end following these steps:
RXXLRXRXL ->
XRXLRXRXL ->
XRLXRXRXL ->
XRLXXRRXL ->
XRLXXRRLX

Note:
1 <= len(start) = len(end) <= 10000.
Both start and end will only consist of characters in {‘L’, ‘R’, ‘X’}.

思路：start里的R可以先出现，L只能后出现。因此iteration的时候，统计R和L的值，如果i位置时，Rs < Re || Ls > Le，则返回false. 同时用两个stringbuilder记录两个字符串，如果相同则返回true，否则返回false.

    public boolean canTransform(String start, String end) {
        if(start.length() != end.length()) return false;
        int r = 0;
        int l = 0;
        StringBuilder s = new StringBuilder();
        StringBuilder e = new StringBuilder();
        for (int i = 0; i < start.length(); i++) {
            char ss = start.charAt(i);
            char ee = end.charAt(i);
            if (ss == 'R') {
                r++;
                s.append(ss);
            } else if (ss == 'L') {
                l++;
                s.append(ss);
            } else {}
            if (ee == 'R') {
                r--;
                e.append(ee);
            } else if (ee == 'L') {
                l--;
                e.append(ee);
            } else {}
            if (r < 0 || l > 0) {
                return false;
            }
        }
        return s.toString().equals(e.toString()) ? true : false;
    }