Description
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:
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| Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".
|
Example 2:
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| Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".
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Example 3:
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| Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".
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Example 4:
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| Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".
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Note:
1 <= S.length <= 2001 <= T.length <= 200S and T only contain lowercase letters and '#' characters.
Follow up:
- Can you solve it in
O(N) time and O(1)space?
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| class Solution {
public:
bool backspaceCompare(string S, string T) {
vector<char> s1, t1;
for(int i = 0;i<S.size();i++)
{
if(S[i] == '#' && !s1.empty())
s1.pop_back();
else if(S[i] == '#' && s1.empty())
continue;
else
s1.push_back(S[i]);
}
for(int i = 0;i<T.size();i++)
{
if(T[i] == '#' && !t1.empty())
t1.pop_back();
else if(T[i] == '#' && t1.empty())
continue;
else
t1.push_back(T[i]);
}
return s1 == t1;
}
};
|
