lintcode(643)System Longest File Path_system longest file path lintcode-优快云博客

本文介绍了一种使用栈来解决寻找文件系统中最长绝对路径的方法。通过解析输入字符串并跟踪每级目录的长度，该算法能在O(n)的时间复杂度内找到最长路径。

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描述：

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.

注意事项

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.

样例：

Give input = "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext" return 20

思路：

使用栈记录每一级目录的长度，当前路径的长度＝当前目录的长度＋上一级路径的长度。当前目录是文件时，统计总的路径长度。所求位最长路径不是最深路径。

public class Solution {
    /**
     * @param input an abstract file system
     * @return return the length of the longest absolute path to file
     */
    public int lengthLongestPath(String input) {
        // Write your code here
        input += '\n';
        char[] temp = input.toCharArray();
        Stack<Integer> result = new Stack<Integer>();
        int len = 0;
        int depth = 0;
        int count = 0;
        boolean isFile = false;
        for(int i = 0; i<temp.length;i++){
            if(temp[i] == '\n'){
                if(depth > 0){
                    count += result.get(depth - 1) + 1;
                }
                while(result.size() > depth){
                    result.pop();
                }
                result.push(count);
                if(isFile){
                    len = Math.max(count , len);
                }
                count = 0;
                depth = 0;
                isFile = false;
            }else if(temp[i] == '\t'){
                depth++;
            }else{
                count++;
                if(temp[i] == '.'){
                    isFile = true;
                }
            }
        }
        return len;
    }
}