Simplify Path

最新推荐文章于 2025-08-14 22:07:01 发布

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#Simplify Path #string #leetcode #算法

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本文介绍了一个使用C++实现的简化Unix风格文件路径的方法。通过栈的数据结构处理多种特殊情况，如多个连续的斜杠、当前目录(.)和父目录(..)等。

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题目

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".

思路

特殊情况的考虑比较麻烦，细心一点就好了。

用栈来操作。

会出现的特殊情况：

"/" ---------------> "/"

"///" ------------------->"/"

"/../" ------------------->"/"

"/.hidden" ---------------->"/.hidden"

"/."------------------------> "/"

"/.." ---------------------->"/"

"/.hidden/" ------------------------->"/.hidden"

"/a/./b/../../c/" ------------------------>"/c"

代码吐下：

class Solution {
public:
    string simplifyPath(string path) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        stack<string> mypath;
        string str = "";
        int len = path.length();
        if(len<=1)
            return path;
        for(int i=0;i<len;i++){
            if(path[i]=='/'){
                if(i>0 && str.length()>0)
                {
                    if(str==".."){
                        if(!mypath.empty()){
                            mypath.pop();
                        }
                    }else if(str!="."){
                        mypath.push(str);
                    }
                }
                str = "";
            }    
            else
                str+=path[i];
        }
        
        if(str==".."){
            if(!mypath.empty()){
                mypath.pop();
            }
        }else if(str!="." && str!=""){
            mypath.push(str);
        }
        str = "";
        if(mypath.empty())
            return "/";
        while(!mypath.empty())
        {
            str = mypath.top()+str;
            mypath.pop();
            str.insert(str.begin(),1,'/');            
        }
        return str;
    }
};