FilesExpand file tree

 master

/

Backspace String Compare.java

Copy path

Blame

More file actions

Blame

More file actions

Latest commit

 

History

History

History

69 lines (57 loc) · 1.43 KB

 master

/

Backspace String Compare.java

Copy path

Top

File metadata and controls

• Code
• Blame

69 lines (57 loc) · 1.43 KB

Raw

Copy raw file

Download raw file

Open symbols panel

Edit and raw actions

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();

}

}

```