The directory dir contains two sub-directories subdir1 and subdir2. subdir1 contains a file file1.ext and an empty second-level sub-directory subsubdir1. subdir2contains a second-level sub-directory subsubdir2 containing a file file2.ext.

We are interested in finding the longest (number of characters) absolute path to a file within our file system. For example, in the second example above, the longest absolute path is"dir/subdir2/subsubdir2/file2.ext", and its length is 32 (not including the double quotes).

Given a string representing the file system in the above format, return the length of the longest absolute path to file in the abstracted file system. If there is no file in the system, return 0.

Note:

• The name of a file contains at least a . and an extension.
• The name of a directory or sub-directory will not contain a ..

Time complexity required: O(n) where n is the size of the input string.

Notice that a/aa/aaa/file1.txt is not the longest file path, if there is another path aaaaaaaaaaaaaaaaaaaaa/sth.png.

思路：将input根据‘\n‘分割，根据开始的‘\t‘数量判断深度。用栈来统计当前的路径长度。

这个题学到了两个点：1是如何分割字符串，2是如何判断一个字符串是否含有某个子串。

 1 class Solution { 2 public: 3     vector<string> split(istringstream& input, char delim) { 4         string item; 5         vector<string> res; 6         while (std::getline(input, item, delim)) { 7             res.push_back(item); 8         } 9         return res;10     }11     int countTb(string path) {12         int res = 1;13         for (int i = 0, n = path.size(); path[i] == ‘\t‘ && i < n; i++)14             res++;15         return res;16     }17     int lengthLongestPath(string input) {18         std::istringstream iss(input);19         vector<string> paths = split(iss, ‘\n‘);20         stack<int> res;21         int maxLen = 0;22         for (string path : paths) {23             int level = countTb(path);24             int lengthInc = path.size() - level + 2;25             while (res.size() >= level) res.pop();26             if (res.empty()) res.push(lengthInc - 1);27             else res.push(res.top() + lengthInc);28             if (path.find(‘.‘) != std::string::npos)29                 maxLen = std::max(maxLen, res.top());30         }31         return maxLen;32     }33 };