Leetcode 71. Simplify Path

https://leetcode.com/problems/simplify-path/

Before doing this problem, you have to know about unix-style. The '. .' means going back one level, and '.' means doing nothing, and others mean going deep.

Using stack operation is a good way to solve. Or we can attack from string operation.

Solution 1:

# T:O(n) S:O(n)
class Solution:
    # @param path, a string
    # @return a string
    def simplifyPath(self, path):
        stack, tokens = [], path.split("/")
        for token in tokens:
            if token == ".." and stack:
                stack.pop()
            elif token != ".." and token != "." and token:
                stack.append(token)
        return "/" + "/".join(stack)

Run Time: 48 ms

Solution 2:

# T:O(n) S:O(n)
class Solution:
    # @param {string} path
    # @return {string}
    def simplifyPath(self, path):
        path = path.split('/')
        cur = '/'
        for i in path:
            if i == '..':  
            # back to upper level
                if cur != '/':
                    cur = '/'.join(cur.split('/')[:-1])  
         # if upper level exist, roll back by one level
                    if cur == '':
                        cur = '/'
            elif i != '.' and i != '':  # add one level
                if cur == '/':
                    cur += i
                else:
                    cur += '/' + i
        return cur

Run Time: 60 ms