[LeetCode] 844. Backspace String Compare 退格字符串比较
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 <= S.length <= 2001 <= T.length <= 200SandTonly contain lowercase letters and'#'characters.
Follow up:
- Can you solve it in
O(N)time andO(1)space?
给2个字符串S和T，里面含有#代表退格键，意味着前面的字符会被删除，判断2个字符串是否相等，字符串中只含有小写字母和#。
解法1：用一个栈存字符，循环字符串，如果是字母就加入栈，遇到#而且栈里面有字母就pop出栈里最后的字符。T: O(n), S:(n)
解法2：follow up要求O(N) time and O(1) space，在一个循环里，从后往前处理字符串，用一个变量记录要删除的字符数量，遇到#时变量加1，遇到字符并且变量大于1，变量减1，直到没遇到#并且要变量为0，这时比较两个字符此时是否一样，不一样返回false，如果字符串比较完没有不一样的字符出现，返回ture。
G家：follow up: 如果有大写键CAP
Java: O(1) space
public boolean backspaceCompare(String S, String T) {
int i = S.length() - 1, j = T.length() - 1;
while (true) {
for (int back = 0; i >= 0 && (back > 0 || S.charAt(i) == '#'); --i)
back += S.charAt(i) == '#' ? 1 : -1;
for (int back = 0; j >= 0 && (back > 0 || T.charAt(j) == '#'); --j)
back += T.charAt(j) == '#' ? 1 : -1;
if (i >= 0 && j >= 0 && S.charAt(i) == T.charAt(j)) {
i--; j--;
} else
return i == -1 && j == -1;
}
}
Python:
def backspaceCompare(self, S, T):
i, j = len(S) - 1, len(T) - 1
backS = backT = 0
while True:
while i >= 0 and (backS or S[i] == '#'):
backS += 1 if S[i] == '#' else -1
i -= 1
while j >= 0 and (backT or T[j] == '#'):
backT += 1 if T[j] == '#' else -1
j -= 1
if not (i >= 0 and j >= 0 and S[i] == T[j]):
return i == j == -1
i, j = i - 1, j - 1
Python:
# Time: O(m + n)
# Space: O(1)
import itertools
class Solution(object):
def backspaceCompare(self, S, T):
"""
:type S: str
:type T: str
:rtype: bool
"""
def findNextChar(S):
skip = 0
for i in reversed(xrange(len(S))):
if S[i] == '#':
skip += 1
elif skip:
skip -= 1
else:
yield S[i]
return all(x == y for x, y in
itertools.izip_longest(findNextChar(S), findNextChar(T)))
Python: wo O(n) space
class Solution(object):
def backspaceCompare(self, S, T):
"""
:type S: str
:type T: str
:rtype: bool
"""
s1, s2 = [], []
for i in range(len(S)):
if S[i] == '#' and s1:
s1.pop()
elif S[i] == '#':
continue
else:
s1.append(S[i])
for j in range(len(T)):
if T[j] == '#' and s2:
s2.pop()
elif T[j] == '#':
continue
else:
s2.append(T[j])
return s1 == s2
C++:
bool backspaceCompare(string S, string T) {
int i = S.length() - 1, j = T.length() - 1;
while (1) {
for (int back = 0; i >= 0 && (back || S[i] == '#'); --i)
back += S[i] == '#' ? 1 : -1;
for (int back = 0; j >= 0 && (back || T[j] == '#'); --j)
back += T[j] == '#' ? 1 : -1;
if (i >= 0 && j >= 0 && S[i] == T[j])
i--, j--;
else
return i == -1 && j == -1;
}
}
