Description
Given two strings s and t, return true 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 = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".
Constraints:
1 <= s.length, 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?
Solutions
Solution 1
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29 | class Solution:
def backspaceCompare(self, s: str, t: str) -> bool:
i, j, skip1, skip2 = len(s) - 1, len(t) - 1, 0, 0
while i >= 0 or j >= 0:
while i >= 0:
if s[i] == '#':
skip1 += 1
i -= 1
elif skip1:
skip1 -= 1
i -= 1
else:
break
while j >= 0:
if t[j] == '#':
skip2 += 1
j -= 1
elif skip2:
skip2 -= 1
j -= 1
else:
break
if i >= 0 and j >= 0:
if s[i] != t[j]:
return False
elif i >= 0 or j >= 0:
return False
i, j = i - 1, j - 1
return True
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38 | class Solution {
public boolean backspaceCompare(String s, String t) {
int i = s.length() - 1, j = t.length() - 1;
int skip1 = 0, skip2 = 0;
for (; i >= 0 || j >= 0; --i, --j) {
while (i >= 0) {
if (s.charAt(i) == '#') {
++skip1;
--i;
} else if (skip1 > 0) {
--skip1;
--i;
} else {
break;
}
}
while (j >= 0) {
if (t.charAt(j) == '#') {
++skip2;
--j;
} else if (skip2 > 0) {
--skip2;
--j;
} else {
break;
}
}
if (i >= 0 && j >= 0) {
if (s.charAt(i) != t.charAt(j)) {
return false;
}
} else if (i >= 0 || j >= 0) {
return false;
}
}
return true;
}
}
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34 | class Solution {
public:
bool backspaceCompare(string s, string t) {
int i = s.size() - 1, j = t.size() - 1;
int skip1 = 0, skip2 = 0;
for (; i >= 0 || j >= 0; --i, --j) {
while (i >= 0) {
if (s[i] == '#') {
++skip1;
--i;
} else if (skip1) {
--skip1;
--i;
} else
break;
}
while (j >= 0) {
if (t[j] == '#') {
++skip2;
--j;
} else if (skip2) {
--skip2;
--j;
} else
break;
}
if (i >= 0 && j >= 0) {
if (s[i] != t[j]) return false;
} else if (i >= 0 || j >= 0)
return false;
}
return true;
}
};
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36 | func backspaceCompare(s string, t string) bool {
i, j := len(s)-1, len(t)-1
skip1, skip2 := 0, 0
for ; i >= 0 || j >= 0; i, j = i-1, j-1 {
for i >= 0 {
if s[i] == '#' {
skip1++
i--
} else if skip1 > 0 {
skip1--
i--
} else {
break
}
}
for j >= 0 {
if t[j] == '#' {
skip2++
j--
} else if skip2 > 0 {
skip2--
j--
} else {
break
}
}
if i >= 0 && j >= 0 {
if s[i] != t[j] {
return false
}
} else if i >= 0 || j >= 0 {
return false
}
}
return true
}
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34 | function backspaceCompare(s: string, t: string): boolean {
let i = s.length - 1;
let j = t.length - 1;
while (i >= 0 || j >= 0) {
let skip = 0;
while (i >= 0) {
if (s[i] === '#') {
skip++;
} else if (skip !== 0) {
skip--;
} else {
break;
}
i--;
}
skip = 0;
while (j >= 0) {
if (t[j] === '#') {
skip++;
} else if (skip !== 0) {
skip--;
} else {
break;
}
j--;
}
if (s[i] !== t[j]) {
return false;
}
i--;
j--;
}
return true;
}
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42 | impl Solution {
pub fn backspace_compare(s: String, t: String) -> bool {
let (s, t) = (s.as_bytes(), t.as_bytes());
let (mut i, mut j) = (s.len(), t.len());
while i != 0 || j != 0 {
let mut skip = 0;
while i != 0 {
if s[i - 1] == b'#' {
skip += 1;
} else if skip != 0 {
skip -= 1;
} else {
break;
}
i -= 1;
}
skip = 0;
while j != 0 {
if t[j - 1] == b'#' {
skip += 1;
} else if skip != 0 {
skip -= 1;
} else {
break;
}
j -= 1;
}
if i == 0 && j == 0 {
break;
}
if i == 0 || j == 0 {
return false;
}
if s[i - 1] != t[j - 1] {
return false;
}
i -= 1;
j -= 1;
}
true
}
}
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