1. 题目

2. 分析

3. 代码

我写了一版很复杂的代码：

class Solution:
    def simplifyPath(self, path: str) -> str:
        operator = [] # 操作符的栈
        dir_name = [] # 文件名的栈
        idx = 0
        cur_dir_name = ""
        while(idx < len(path)):
            if path[idx] == '/':
                operator.append('/')
                cur_dir_name = "" # init
                next_idx = idx+1
                while(next_idx < len(path) and path[next_idx]!='/'):
                    cur_dir_name += path[next_idx]
                    next_idx+=1
                # 获取dir_name
                if cur_dir_name == ".":
                    operator.pop()
                elif cur_dir_name == "..":
                    if len(operator):
                        operator.pop()
                    if len(dir_name):
                        dir_name.pop()
                elif cur_dir_name != "":
                    dir_name.append(cur_dir_name)
                elif cur_dir_name == "":
                    operator.pop()
                idx = next_idx
        # 输出最后结果
        res = ""
        for i in range(len(operator)):
            if i < len(dir_name):
                res += (operator[i] + dir_name[i])
        if res == "":
            res = "/"
        return res