LeetCode #844 BackspaceStringCompare. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,그래프,Graph,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접, 데이터베이스, sql, query, 쿼리
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Backspace String Compare.
Memory Usage: 8.4 MB, less than 81.82% of C++ online submissions for Backspace String Compare.
LeetCode #844
Q.
Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.
주어진 문자열 S, T 가 있는데, 빈 텍스트 에디터에 입력되었을 때 같은 문자인지 확인해서 리턴해라. # 은 백스페이스를 의미하는 문자이다.
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.
Process
// Process
//1. Input two strings S, T
//2. Iterate from begin to the end
// 2.1. Check if it's # or not
// 2.1.1. If so -> delete front char
// 2.2. Delete # char
//3. Return result after checking they are same or not
// 처리과정
//1. 문자열 S, T 입력받는다.
//2. 시작부터 끝까지 반복한다.
// 2.1. 해당 문자가 # 인지 확인해서
// 2.1.1. # 이면 -> 앞의 글자를 지운다
// 2.2. # 글자 지운다
//3. 같은지 확인해서 결과 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
bool backspaceCompare(string S, string T) {
bool result = false;
for (int i = 0; i < S.length(); ++i) {
if (S[i] == '#') {
if (i > 0) {
S.erase(S.begin() + (i-1));
--i;
}
S.erase(S.begin() + i);
--i;
}
}
for (int i = 0; i < T.length(); ++i) {
if (T[i] == '#') {
if (i > 0) {
T.erase(T.begin() + (i-1));
--i;
}
T.erase(T.begin() + i);
--i;
}
}
if (S == T) {
result = true;
}
return result;
}
};
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