LeetCode #71 - Simplify Path

最新推荐文章于 2020-09-18 20:20:43 发布

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本文介绍了一个算法问题，即如何简化Unix风格的文件路径。通过分析和处理包含'.','..'及冗余'/'的路径，提供了C++代码实现，展示了如何将复杂的路径转换为最简形式。

题目描述：

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

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

In a UNIX-style file system, a period ('.') refers to the current directory, so it can be ignored in a simplified path. Additionally, a double period ("..") moves up a directory, so it cancels out whatever the last directory was. For more information, look here: https://en.wikipedia.org/wiki/Path_(computing)#Unix_style

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".

将字符串按照’/’拆分，每一部分分情况处理，并保存起来。

class Solution {
public:
    string simplifyPath(string path) {
        string s;
        vector<string> v;
        istringstream iss(path);
        while(getline(iss,s,'/'))
        {
            if(s.empty()||s==".") continue;
            else if(s=="..")
            {
                if(v.size()==0) continue;
                else v.pop_back();
            }
            else v.push_back(s);
        }
        string result;
        for(string x:v) result+="/"+x;
        if(result=="") result="/";
        return result;
    }
};