-
-
Notifications
You must be signed in to change notification settings - Fork 31
Expand file tree
/
Copy pathBackspace String Compare.java
More file actions
executable file
·69 lines (57 loc) · 1.43 KB
/
Copy pathBackspace String Compare.java
File metadata and controls
executable file
·69 lines (57 loc) · 1.43 KB
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
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
E
1533516441
tags: Two Pointers, Stack
```
/*
Given two strings S and T, return if they are equal when both are typed into empty text editors.
# means a backspace character.
Example 1:
Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".
Example 2:
Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".
Example 3:
Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".
Example 4:
Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".
Note:
1 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and '#' characters.
Follow up:
Can you solve it in O(N) time and O(1) space?
*/
/*
Stack to take elements, and pop when # if reached.
compare the two stacks at the end.
time/space: O(n)
*/
class Solution {
public boolean backspaceCompare(String S, String T) {
if (S == null || T == null) return false;
return type(S).equals(type(T));
}
private String type(String s) {
Stack<Character> stack = new Stack<>();
for (char c : s.toCharArray()) {
if (c == '#') {
if (!stack.isEmpty()) stack.pop();
} else {
stack.push(c);
}
}
StringBuffer sb = new StringBuffer();
while (!stack.isEmpty()) {
sb.append(stack.pop());
}
return sb.toString();
}
}
```