力扣 在LR字符串中交换相邻字符(双指针)

题目：传送门

思路：记录’X‘值之外字符的坐标（或者双指针）

代码：
c++（记录坐标）

class Solution {
public:
    vector<node> v, v1;
    bool canTransform(string s, string e) {
        for(int i = 0; i < s.length(); i++) {
            if(s[i] == 'X')continue;
            v.push_back(node(i, s[i]));
        }
        for(int i = 0; i < e.length(); i++) {
            if(e[i] == 'X')continue;
            v1.push_back(node(i, e[i]));
        }
        if(v1.size() != v.size())return false;
        for(int i = 0; i <v1.size(); i++) {
            if(v1[i].c != v[i].c)return false;
            else if(v1[i].c == 'R' && v1[i].p < v[i].p) {
                return false;
            }
            else if(v1[i].c == 'L' && v1[i].p > v[i].p) {
                return false;
            }
        }
        return true;
    }
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

python(双指针)

class Solution:
    def canTransform(self, s: str, e: str) -> bool:
        i = 0
        j = 0
        n = len(s)
        while i < n and j < n:
            while i < n and s[i] == 'X':
                i += 1
            while j < n and e[j] == 'X':
                j += 1
            if i >= n or j >= n:
                continue
            if s[i] == e[j]:
                if s[i] == 'L' and i < j:
                    return False
                if s[i] == 'R' and i > j:
                    return False
            else:
                return False
            i += 1
            j += 1
        while i < n and s[i] == 'X':
            i += 1
        while j < n and e[j] == 'X':
            j += 1  
        if i < n or j < n:
            return False
        return True
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28