Given two strings S
and T
, return 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 = "#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
andT
only contain lowercase letters and'#'
characters.
Follow up:
- Can you solve it in
O(N)
time andO(1)
space?
----------------------
Several points should be ready for O(n) and O(1) solution:
1. Find next un-skipped character reversely.
2. Use skip variable to record the skipping num. If else in 3 branches
3. Take care one string terminates but the other one not:
class Solution:
def backspaceCompare(self, S: str, T: str) -> bool:
def get_pre_pos(s,i):
skip = 0
while (i>=0):
if (s[i] == '#'):
skip,i = skip+1,i-1
elif (skip > 0):
skip,i = skip-1,i-1
else:
return i
return i
i,j = len(S)-1,len(T)-1
while (True):
i,j = get_pre_pos(S,i),get_pre_pos(T,j)
#print("i={0} j={1}".format(i,j))
if (i>=0 and j>=0 and S[i] == T[j]):
i,j=i-1,j-1
elif (i<0 and j<0):
return True
else:
return False