我报名参加金石计划1期挑战——瓜分10万奖池，这是我的第4篇文章，点击查看活动详情
描述
You are given a string s, which contains stars *. In one operation, you can:
- Choose a star in s.
- Remove the closest non-star character to its left, as well as remove the star itself.
Return the string after all stars have been removed.
Note:
- The input will be generated such that the operation is always possible.
- It can be shown that the resulting string will always be unique.
Example 1:
Input: s = "leet**cod*e"
Output: "lecoe"
Explanation: Performing the removals from left to right:
- The closest character to the 1st star is 't' in "leet**cod*e". s becomes "lee*cod*e".
- The closest character to the 2nd star is 'e' in "lee*cod*e". s becomes "lecod*e".
- The closest character to the 3rd star is 'd' in "lecod*e". s becomes "lecoe".
There are no more stars, so we return "lecoe".
Example 2:
Input: s = "erase*****"
Output: ""
Explanation: The entire string is removed, so we return an empty string.
Note:
1 <= s.length <= 10^5
s consists of lowercase English letters and stars *.
The operation above can be performed on s.
解析
根据题意，给定一个字符串 s ，其中包含星号 * 。 在一次操作中，我们可以：
- 在 s 中选择一颗星号
- 移除最靠近其左侧的非星形字符，以及移除星形本身
移除所有星号后返回字符串。
注意：
- 生成的输入保证始终可以进行上述操作
- 可以证明生成的字符串总是唯一的
其实这道题很简单，主要考察的是数据结构——栈的使用，我们直接遍历字符串 s ，当字符不为星号的时候，我们将该字符拼接到结果 result 后面，如果当前的字符为星号的时候，如果现有的 result 不为空，我们就将最后一个字符弹出去掉，这样遍历结束之后即可得到最后的结果 result ，直接返回 result 即可。
时间复杂度为 O(N) ，空间复杂度为 O(N) 。
解答
class Solution(object):
def removeStars(self, s):
"""
:type s: str
:rtype: str
"""
result = []
s = list(s)
for c in s:
if c == '*':
if result:
result.pop()
else:
result.append(c)
return ''.join(result)
运行结果
65 / 65 test cases passed.
Status: Accepted
Runtime: 637 ms
Memory Usage: 17.1 MB
原题链接
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