leetcode 388. Longest Absolute File Path 最长长度文件路径 + 广度优先遍历BFS

墨蓝 2022-06-04 05:57 128阅读 0赞

Suppose we abstract our file system by a string in the following manner:

The string “dir\\n\\tsubdir1\\n\\tsubdir2\\n\\t\\tfile.ext” represents:

dir  
subdir1  
subdir2  
file.ext  
The directory dir contains an empty sub-directory subdir1 and a sub-directory subdir2 containing a file file.ext.

The string “dir\\n\\tsubdir1\\n\\t\\tfile1.ext\\n\\t\\tsubsubdir1\\n\\tsubdir2\\n\\t\\tsubsubdir2\\n\\t\\t\\tfile2.ext” represents:

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

本题就是做一个BFS广度优先遍历来做，网上的答案很棒，很值得学习，

提示：可以使用**map【string,set【string】】来保存树的结果**

需要注意的地方时，本题的做法很棒，外部的for循环做遍历，内部有while遍历，这样刚好把”\\n”跳过去了，注意观察i的变化。

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    
    using namespace std;
    
    class Solution
    {
    public:
        int lengthLongestPath(string input) 
        {
            int maxLen = 0, count = 0, lev = 1;
            bool isFile = false;
            vector<int> level(input.size()+1,0);
            for(int i = 0; i < input.size();i++)
            {
                while (input[i] == '\t')
                {
                    lev++;
                    i++;
                }
    
                while (i < input.size() && input[i] != '\n')
                {
                    if (input[i] == '.')
                        isFile = true;
                    i++;
                    count++;
                }
    
                if (isFile)
                    maxLen = max(maxLen,level[lev-1]+count);
                else
                    level[lev] = level[lev - 1] + count + 1;
                count = 0;
                lev = 1;
                isFile = false;
            }
            return maxLen;
        }
    };