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

Constraints:

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

思路：發現規律：

    // R 只能右邊移動, R不能跳過L和R
    // L 只能左邊移動；L不能跳過L和R
    // 而且所有R,L去掉X之後的相對順序和個數不能變

 雙指針，掃描到不是X的地方，判斷相對性，R和L一定要相同，而且座標，R p2不能小於P1，L，p2不能大於p1

class Solution {
    // R 只能右邊移動, R不能跳過L和R
    // L 只能左邊移動；L不能跳過L和R
    // 而且所有R,L去掉X之後的相對順序和個數不能變
    public boolean canTransform(String start, String end) {
        if(!start.replace("X", "").equals(end.replace("X", ""))) {
            return false;
        }
        if(start.length() != end.length()) {
            return false;
        }
        int p1 = 0; int p2 = 0;
        while(p1 < start.length() && p2 < end.length()) {
            // get Non X position;
            while(p1 < start.length() && start.charAt(p1) == 'X') {
                p1++;
            }
            while(p2 < end.length() && end.charAt(p2) == 'X') {
                p2++;
            }
            
            if(p1 == start.length() && p2 == end.length()) {
                return true;
            }
            
            if(p1 == start.length() || p2 == end.length()) {
                return false;
            }
            
            if(start.charAt(p1) != end.charAt(p2)) {
                return false;
            }
            
            if(start.charAt(p1) =='L' && p2 > p1) {
                return false;
            }
            if(start.charAt(p1) == 'R' && p2 < p1) {
                return false;
            }
            p1++;
            p2++;
        }
        return true;
    }
}