LintCode 421 Simplify Path (字符串处理题)

421 · Simplify Path
Algorithms

Description
Given an absolute path for a file (Unix-style), simplify it.

In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period … moves the directory up a level.

The result must always begin with /, and there must be only a single / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the result must be the shortest string representing the absolute path.

Did you consider the case where path is “/…/”?

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”.

Example
Example 1:

Input: “/home/”
Output: “/home”
Example 2:

Input: “/a/./…/…/c/”
Output: “/c”
Explanation: “/” has no parent directory, so “/…/” equals “/”.
Tags
Company
OpenAI
Facebook
Microsoft

class Solution {
public:
    /**
     * @param path: the original path
     * @return: the simplified path
     */
    string simplifyPath(string &path) {
        int n = path.size();
        string res = "";
        int pos = 0, dirStartPos = 0, dirEndPos = 0;
        stack<string> stk;
        while (pos < n) {
            while (pos < n && path[pos] == '/') pos++;
            dirStartPos = pos;
            while (pos < n && path[pos] != '/') pos++;
            dirEndPos = pos;
            string dirStr = path.substr(dirStartPos, dirEndPos - dirStartPos);
            if (dirStr == "..") {
                if (!stk.empty()) stk.pop();
            } else if (dirStr.size() > 0 && dirStr != ".") {
                stk.push(dirStr);
            }
        }
        while (!stk.empty()) {
            res = "/" + stk.top() + res;
            stk.pop(); 
        }
        if (res.size() == 0) res = "/";
        return res;
    }
};

不用stack用vector也可以。

class Solution {
public:
    /**
     * @param path: the original path
     * @return: the simplified path
     */
    string simplifyPath(string &path) {
        vector<string> dirs;
        int pos = 0;
        while (pos < path.size()) {
            while (path[pos] == '/' && pos < path.size()) ++pos;
            if (pos == path.size()) break;
            int startPos = pos;
            while (path[pos] != '/' && pos < path.size()) ++pos;
            int endPos = pos - 1;
            string s = path.substr(startPos, endPos - startPos + 1);
            if (s == "..") {
                if (!dirs.empty()) dirs.pop_back();
            } else if (s != ".") {
                dirs.push_back(s);
            }
        }

        if (dirs.empty()) return "/";
        string result;
        for (int i = 0; i < dirs.size(); ++i) {
            result += '/' + dirs[i];
        }
    
        return result;
    }
};