【LeetCode】844. Backspace String Compare 解题报告(Python)
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址:https://leetcode.com/problems/backspace-string-compare/description/
题目描述
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 <= 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?
题目大意
在一个空白的编辑器里连续输入两段字符,其中#代表退格,要求最后两段字符是否相同。
有个Follow up,问我们能不能使用O(n)的时间复杂度和O(1)的空间复杂度。
解题方法
字符串切片
字符串题对于Python而言都不算题。就是按照题目要求做一遍就好了。
遇到#,字符串不为空,就删除最后一个字符。如果不是#号,就拼接到字符串的最后。把两个字符串都求出来,然后比较就好。
注意,我不小心踏进了一个坑,因为看到两个连续的if,就把它们合并在一起了,其实不行的:
if s == '#':
if ans_S:
ans_S = ans_S[:-1]
我给改成了:
if s == '#' and ans_S:
ans_S = ans_S[:-1]
这样看着好看了,其实是错的。因为如果字符串是空的,那么输入#号,会把这个#号拼接到字符串上去。
Follow up的要求暂时不会。
代码如下:
class Solution(object):
def backspaceCompare(self, S, T):
"""
:type S: str
:type T: str
:rtype: bool
"""
ans_S = ""
ans_T = ""
for s in S:
if s == '#':
if ans_S:
ans_S = ans_S[:-1]
else:
ans_S += s
for t in T:
if t == '#':
if ans_T:
ans_T = ans_T[:-1]
else:
ans_T += t
return ans_S == ans_T
栈
使用一个栈的话,可以完美处理这个问题,遇到#退栈就好了,唯一需要注意的时候如果栈是空的时候,不能退栈。
class Solution(object):
def backspaceCompare(self, S, T):
"""
:type S: str
:type T: str
:rtype: bool
"""
stackS, stackT = [], []
for s in S:
if s != "#":
stackS.append(s)
elif stackS:
stackS.pop()
for t in T:
if t != "#":
stackT.append(t)
elif stackT:
stackT.pop()
return stackS == stackT
日期
2018 年 6 月 10 日 —— 等了两天的腾讯比赛复赛B的数据集,结果人家在复赛刚开始就给了。。
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