Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.
In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix
Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.
Example 1:
Input: "/home/" Output: "/home" Explanation: Note that there is no trailing slash after the last directory name.
Example 2:
Input: "/../" Output: "/" Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.
Example 3:
Input: "/home//foo/" Output: "/home/foo" Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
Example 4:
Input: "/a/./b/../../c/" Output: "/c"
Example 5:
Input: "/a/../../b/../c//.//" Output: "/c"
Example 6:
Input: "/a//b////c/d//././/.." Output: "/a/b/c"
解題思路：將絕對路徑轉化爲最簡的相對路徑
關於‘/’處理：我們需要關注的是斜槓分隔開的每個文件名，所以遇到斜槓直接遍歷到下一個非斜槓的位置，記錄每個文件夾名。
關於".."處理：這是返回上一級文件夾，所以要將我們之前訪問到的文件夾名刪除（如果存在的話）
關於"."處理：指當前文件夾，無需處理，也不要存儲。
class Solution {
public:
string simplifyPath(string path) {
vector<string>ans;
string res = "";
int i = 0;
while(i < path.length()){
string temp = "";
while(i < path.length() && path[i] == '/')
++ i;
while(i < path.length() && path[i] != '/')
temp += path[i++];
if(temp == ".."){
if(!ans.empty())
ans.pop_back();
}
else if(temp != "."&& temp != "")
ans.push_back(temp);
}
if(ans.empty())
res = "/";
for(int i = 0; i < ans.size(); ++ i)
res += '/' + ans[i];
return res;
}
};