題目描述

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

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

本題要求對給定的Unix路徑進行簡化，關鍵點在於理解Unix路徑的規則以及對一些邊界條件要加以考慮。

Unix路徑規則：

Corner Cases:
path = “/../” => 輸出 “/”.
連續的多個分隔符 ‘/’ , 例如 “/home//foo/” => 忽略重複的分隔符，輸出 “/home/foo”.

解題思路

總體來說本題並不難，只要根據unix path規則，細心考慮可能的邊界條件即可。

class Solution {
public:
    string simplifyPath(string path) {
        char slash = '/';
        vector<string> dirs;
        int begin = 0, end = 0;
        int len = path.length();
        while (end < len)
        {
            begin = end
 
;
            while (begin < len && path[begin] == slash)
                begin++;
            end = begin;
            while (end < len && path[end] != slash)
                end++;
            if (begin != end)
            {
                string tmp(path, begin, end - begin);
                if (tmp == ".." && !dirs.empty())
                    dirs.pop_back();
                else if (tmp == ".." && dirs.empty())
                    continue;
                else if (tmp != ".")
                    dirs.push_back(tmp);
            }

        }

        string simplePath = "";
        for (size_t i = 0; i != dirs.size(); ++i)
        {
            simplePath += slash + dirs[i];
        }

        if(dirs.empty())
            simplePath = slash;
        return simplePath;
    }
};