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".
这道题让简化给定的路径，光根据题目中给的那一个例子还真不太好总结出规律，应该再加上两个例子 path = "/a/./b/../c/", => "/a/c"和path = "/a/./b/c/", => "/a/b/c"， 这样我们就可以知道中间是"."的情况直接去掉，是".."时删掉它上面挨着的一个路径，而下面的边界条件给的一些情况中可以得知，如果是空的话返回"/"，如果有多个"/"只保留一个。那么我们可以把路径看做是由一个或多个"/"分割开的众多子字符串，把它们分别提取出来一一处理即可，代码如下：
C++ 解法一:
class Solution { public: string simplifyPath(string path) { vector<string> v; int i = 0; while (i < path.size()) { while (path[i] == '/' && i < path.size()) ++i; if (i == path.size()) break; int start = i; while (path[i] != '/' && i < path.size()) ++i; int end = i - 1; string s = path.substr(start, end - start + 1); if (s == "..") { if (!v.empty()) v.pop_back(); } else if (s != ".") { v.push_back(s); } } if (v.empty()) return "/"; string res; for (int i = 0; i < v.size(); ++i) { res += '/' + v[i]; } return res; } };
还有一种解法是利用了C语言中的函数strtok来分隔字符串，但是需要把string和char*类型相互转换，转换方法请猛戳这里。除了这块不同，其余的思想和上面那种解法相同，代码如下：
C 解法一:
class Solution { public: string simplifyPath(string path) { vector<string> v; char *cstr = new char[path.length() + 1]; strcpy(cstr, path.c_str()); char *pch = strtok(cstr, "/"); while (pch != NULL) { string p = string(pch); if (p == "..") { if (!v.empty()) v.pop_back(); } else if (p != ".") { v.push_back(p); } pch = strtok(NULL, "/"); } if (v.empty()) return "/"; string res; for (int i = 0; i < v.size(); ++i) { res += '/' + v[i]; } return res; } };
C++中也有专门处理字符串的机制，我们可以使用stringstream来分隔字符串，然后对每一段分别处理，思路和上面的方法相似，参见代码如下：
C++ 解法二：
class Solution { public: string simplifyPath(string path) { string res, t; stringstream ss(path); vector<string> v; while (getline(ss, t, '/')) { if (t == "" || t == ".") continue; if (t == ".." && !v.empty()) v.pop_back(); else if (t != "..") v.push_back(t); } for (string s : v) res += "/" + s; return res.empty() ? "/" : res; } };
Java 解法二：
public class Solution { public String simplifyPath(String path) { Stack<String> s = new Stack<>(); String[] p = path.split("/"); for (String t : p) { if (!s.isEmpty() && t.equals("..")) { s.pop(); } else if (!t.equals(".") && !t.equals("") && !t.equals("..")) { s.push(t); } } List<String> list = new ArrayList(s); return "/" + String.join("/", list); } }
本文转自博客园Grandyang的博客，原文链接：简化路径[LeetCode] Simplify Path ，如需转载请自行联系原博主。
