Leetcode Simplify Path

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"

.

题目不难，主要考虑一些特殊情况。

对于path =

"/a/./b/../../c/"

, =>

"/c"，模拟一下

先按照'/'对字符串进行分割，得到 [a, . , b, .. , .. , c]

首先进入目录a，注意 '.' 代表当前目录 ，".."代表上一个目录

然后到达'.'，还是在当前目录，/a

然后到达'b'，这为/a/b

然后到达'..'，这是回到父目录，则变为/a

然后到达'..'，继续回到父目录，则变为/

然后到达'c'，则达到子目录，变为/c

class Solution {
public:
vector<string> split(string& path, char ch){
int index = 0;
vector<string> res;
while(index < path.length()){
while(index < path.length() && path[index] == '/') index++;
if(index >= path.length()) break;
int start=index, len = 0;
while(index < path.length() && path[index]!='/') {index++;len++;}
res.push_back(path.substr(start,len));
}
return res;
}

string simplifyPath(string path) {
vector<string> a = split(path,'/');
vector<string> file;
for(int i = 0 ; i < a.size(); ++ i){
if(a[i] == ".." ){
if(!file.empty()) file.pop_back();
}
else if(a[i]!=".") file.push_back(a[i]);
}
string res="";
if(file.empty()) res ="/";
else{
for(int i = 0 ; i < file.size(); ++ i) res+="/"+file[i];
}
return res;
}
};