Leetcode 71 Simplify Path

最新推荐文章于 2020-12-19 05:33:40 发布

原创最新推荐文章于 2020-12-19 05:33:40 发布 · 291 阅读

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本文深入探讨了Unix风格文件系统的路径简化算法，通过实例解析如何处理绝对路径中的符号.和..，并提供了详细的代码实现。文章还讨论了特殊情况，如路径中出现连续的斜杠以及/../的情况，确保读者能够全面理解并掌握路径简化的技巧。

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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".

这个题的意思是仿照Unix的路径规划来进行模拟

class Solution {
public:
    string simplifyPath(string path) {
        path += '/';//首先加上'/'以简化操作
        string res,s;
        for(auto c : path){
            if(res.empty()){//res为实际的串
                res += c;
            }else if(c != '/'){//读入正常的串（两个'/'之间的串）
                s += c;
            }else{
                if(s == ".."){
                    if(res.size() > 1){
                        res.pop_back();
                    }
                    while(res.back() != '/'){//连续退,主要是在两个'/'为一段很长的串的时候
                        res.pop_back();
                    }
                }else if(s != "" && s != "."){//正常添加
                    res += s + '/';
                }
                s = "";//将这个串清零
            }
        }
        if(res.size() > 1){
            res.pop_back();//去掉'/'
        }
        return res;
    }
};