71. Simplify Path

I don't have a clue in first place. But after looking at the tag "stack", then I got an idea. We can store each directory into a stack. Then I'll have following case:

1. directory is empty (i.e. consecutive slashes "///")or directory is ".", then we don't have to do anything.
2. directory is "..", then we need to pop out one directory. Note we need to check the stack is empty or not. If it's empty, we don't have to do anything.
3. Otherwise, we push the directory into the stack.

Since we eventually need to pop out the directories to form a new path but the stack is in reverse order, so we can use a vector to replace the stack so that we can traverse the vector from the beginning to the end.

1:  class Solution {  
2:  public:  
3:    string simplifyPath(string path) {  
4:      vector<string> dirs;  
5:      int i = 0, j = 0;  
6:      while (i < path.size()) {  
7:        while (i < path.size() && path[i] != '/') i++;  
8:        string dir = path.substr(j, i-j);  
9:        j = ++i;  
10:        if (dir == "" || dir == ".") continue;  
11:        if (dir == ".." && !dirs.empty()) dirs.pop_back();  
12:        else if (dir != "..") dirs.push_back(dir);  
13:      }  
14:      string res;  
15:      for (int i = 0; i < dirs.size(); i++) {  
16:        res += "/" + dirs[i];  
17:      }  
18:      return res.empty() ? "/" : res;  
19:    }  
20:  };