844 Backspace String Compare 比较含退格的字符串
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:
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?
题目描述:
给定 S 和 T 两个字符串，当它们分别被输入到空白的文本编辑器后，判断二者是否相等，并返回结果。 # 代表退格字符。
示例 :
示例 1：
输入：S = "ab#c", T = "ad#c"
输出：true
解释：S 和 T 都会变成 “ac”。
示例 2：
输入：S = "ab##", T = "c#d#"
输出：true
解释：S 和 T 都会变成 “”。
示例 3：
输入：S = "a##c", T = "#a#c"
输出：true
解释：S 和 T 都会变成 “c”。
示例 4：
输入：S = "a#c", T = "b"
输出：false
解释：S 会变成 “c”，但 T 仍然是 “b”。
提示：
1 <= S.length <= 200
1 <= T.length <= 200
S 和 T 只含有小写字母以及字符 '#'。
思路:
- 直接在字符串上使用 backspace
时间复杂度O(n), 空间复杂度O(1) - 使用栈记录 backspace的结果
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
bool backspaceCompare(string S, string T)
{
int s_index = backspace(S), t_index = backspace(T);
if (s_index != t_index) return false;
else
{
S = S.substr(0, s_index);
T = T.substr(0, t_index);
if (S == T) return true;
}
return false;
}
private:
int backspace (string& S)
{
int index = 0;
for (auto i : S)
{
if (i != '#') S[index++]=i;
else index == 0 ? 0 : --index;
}
return index;
}
};
Java:
class Solution {
public boolean backspaceCompare(String S, String T) {
Stack s = new Stack<>();
Stack t = new Stack<>();
backspaceString(S, s);
backspaceString(T, t);
if (s.size() != t.size()) return false;
while (s.size() != 0) if (!s.isEmpty() && !t.isEmpty() && s.pop() != t.pop()) return false;
return true;
}
private void backspaceString(String S, Stack s) {
for (Character c : S.toCharArray()) {
if (c == '#' && !s.isEmpty()) s.pop();
else if (c != '#') s.push(c);
}
}
}
Python:
class Solution:
def backspaceCompare(self, S: str, T: str) -> bool:
s, t = [], []
def backspace_string(S: str, s: List[str]) -> None:
for i in range(len(S)):
if S[i] != '#':
s.append(S[i])
elif S[i] == '#' and s:
s.pop()
backspace_string(S, s)
backspace_string(T, t)
return s == t