[Leetcode 60] 71 Simplify Path

Problem:

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".

Analysis:

Simulation problem, with the help of a vector keeping record of the current path and either push a new path node or pop the older path node or do nothing to maintain the final path.

Need to pay special attention to those corner cases.

Code:

class Solution {
public:
    string simplifyPath(string path) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> stack;
        stack.push_back("/");
        
        int p1 = 0, p2 = 0;
        while (true) {
            while (path[p1] == '/' && p1<path.size())
                p1++;
                
            p2 = p1+1;
            while (path[p2] != '/' && p2<path.size()) 
                p2++;  
            
            if (p2 > path.size()) break;
            
            string tmp = path.substr(p1, p2-p1);        
            if (tmp == "..") {
                if (stack.size() != 1)
                     stack.pop_back();
            }
            else if (tmp == ".")
                ;//do nothing
            else
                stack.push_back(tmp);
            
            p1 = p2;
        }
        
        
        if (stack.size() == 1)
            return "/";
        else {
            string tmp;
            for (int i=1; i<stack.size(); i++)
                tmp += "/" + stack[i];
            return tmp;
        }
    }
};