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BackSpaceStringCompare.cpp
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BackSpaceStringCompare.cpp
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/*
844. Backspace String Compare
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. 1 <= S.length <= 200
2. 1 <= T.length <= 200
3. S and T only contain lowercase letters and '#' characters.
Follow up:
• Can you solve it in O(N) time and O(1) space?
From <https://leetcode.com/problems/backspace-string-compare/description/>
*/
class Solution {
public:
bool backspaceCompare(string S, string T) {
stack<char> myStack;
for(int i = 0; i < S.size(); i++) {
if(S[i] != '#') {
myStack.push(S[i]);
} else if(S[i] == '#' && !myStack.empty()) {
myStack.pop();
}
}
string s = "";
while(!myStack.empty()) {
s += myStack.top();
myStack.pop();
}
for(int i = 0; i < T.size(); i++) {
if(T[i] != '#') {
myStack.push(T[i]);
} else if(T[i] == '#' && !myStack.empty()) {
myStack.pop();
}
}
string t = "";
while(!myStack.empty()) {
t += myStack.top();
myStack.pop();
}
return s == t;
}
};