> For the complete documentation index, see [llms.txt](https://r24zeng.gitbook.io/leetcode-notebook/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://r24zeng.gitbook.io/leetcode-notebook/bu-chong-2140-dao/71.-simplify-path.md).

# 71. Simplify Path

\# Medium

{% hint style="info" %}
Key idea:

1. extract the string between start`'/'` and end`'/'` storing in a string list(only effctive path)
2. `"."` do nothing
3. `".."` path2.pop()
   {% endhint %}

### Solution:

1. find the effective start `'/'`
2. find the effective end `'/'`, extract string between start and end
3. if string = `'.'`, do nothing; if string = `'..'`, `path2.pop()`; else `path2.append(string)`
4. convert `path2` list to a real path string

```python
class Solution:
    def simplifyPath(self, path: str) -> str:
        i = 0
        path2 = []
        while i < len(path):
            # find start '/', i == start'/'
            while i < len(path) and path[i] == '/':
                i += 1
            if i == len(path):
                break
                
            # find the end '/', i == end'/'
            tmp = ''
            while i < len(path) and path[i] != '/':
                tmp += path[i]
                i += 1
            
            # add to path2
            if tmp == '..':
                if len(path2) != 0:
                    path2.pop()
            elif tmp == '.':
                continue
            else:
                path2.append(tmp)
                
        # convert path2 to canonical path
        if len(path2) == 0: return '/'
        
        c = ''
        for i in range(len(path2)):
            c = c + '/' + path2[i]

        
        return c
```

Time complexity = $$O(n)$$ , space complexity = $$O(n)$$&#x20;