Leetcode
2020.05.05 12:45

# 844. Backspace String Compare

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크게 작게 위로 아래로 댓글로 가기 인쇄

Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

Note that after backspacing an empty text, the text will continue empty.

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.

• Can you solve it in O(N) time and O(1) space?

class Solution {
public boolean backspaceCompare(String S, String T) {
return makeString(S).equals(makeString(T));
}

public List<Character> makeString(String str){
List<Character> list = new ArrayList<>();

for(int i=0; i<str.length(); i++){
if(str.charAt(i) == '#'){
if(list.size() > 0){
list.remove(list.size() - 1);
}
}else{
}
}

//return list.stream().map(e->e.toString()).collect(Collectors.joining());
return list;
}
}

class Solution {
public boolean backspaceCompare(String S, String T) {
return makeString(S).equals(makeString(T));
}

public Stack<Character> makeString(String str){
Stack<Character> stack = new Stack();

for(int i=0; i<str.length(); i++){
if(str.charAt(i) != '#'){
stack.push(str.charAt(i));
}else if(stack.size() > 0){
stack.pop();
}
}

//return String.valueOf(stack);
return stack;
}
}

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