LeetCode: Simplify Path

最新推荐文章于 2022-01-28 00:45:13 发布

Tingmei

最新推荐文章于 2022-01-28 00:45:13 发布

阅读量1.1k

点赞数

CC 4.0 BY-SA版权

分类专栏： C++ 算法

本文链接：https://blog.youkuaiyun.com/Tingmei/article/details/8054060

算法同时被 2 个专栏收录

109 篇文章

订阅专栏

C++

108 篇文章

订阅专栏

本文介绍了一个C++函数，用于简化Unix风格文件系统的绝对路径。通过解析输入路径并利用栈来跟踪目录层级，该函数能有效去除多余的'.'和'..'符号，并处理连续的斜杠。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

Corner Cases:

Did you consider the case where path = "/../"?
In this case, you should return "/".
Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/".
In this case, you should ignore redundant slashes and return "/home/foo".

class Solution {
public:
    string simplifyPath(string path) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = path.size();
        if (len == 0) return "/";
        vector<string> stk;
        int left = '/' == path[0]? 0: -1;
        int right;
        while(left < len)
        {
            right = left;
            while(++right < len && '/' != path[right]);
            string part = path.substr(left+1, right-left-1);
            left = right;
            if("." == part || "" == part)  
                continue;  
            else if(".." == part)  
            {  
                if(stk.size() > 0)
                    stk.pop_back();  
            }  
            else  
                stk.push_back(part);  
        }
        
        if (stk.size() > 0)
        {
            string result = "";
            for (int i = 0; i < stk.size(); ++i)
            {
                result += '/' + stk[i];
            }
            return result;
        }
        else
            return "/";
    }
};