[LeetCode] 71. Simplify Path 简化路径-优快云博客

本文介绍了一个简化Unix风格文件路径的方法，通过使用多种编程语言（包括Java、Python和C++）实现，来解决路径中出现的多余符号和层级跳转问题。

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".

规律：

1、".."表示跳到上一层目录，删掉它上面挨着的一个路径。

2、"."表示当前目录，直接去掉。

3、如果有多个"/"只保留一个。

4、如果最后有一个"/"，去掉不要。

Java:

public class Solution {
    public String simplifyPath(String path) {
        java.util.LinkedList<String> stack = new java.util.LinkedList<String>();
        java.util.Set<String> set = new java.util.HashSet<String>(java.util.Arrays.asList("..", ".", ""));
        for (String str : path.split("/")){
            if (str.equals("..") && !stack.isEmpty()) stack.pop();
            if (!set.contains(str))    stack.push(str);
        }

        String result = "";
        for (String str : stack)  result =  "/" + str + result;
        return result.isEmpty() ? "/" : result;
    }
}

Python:

class Solution:
    # @param path, a string
    # @return a string
    def simplifyPath(self, path):
        stack, tokens = [], path.split("/")
        for token in tokens:
            if token == ".." and stack:
                stack.pop()
            elif token != ".." and token != "." and token:
                stack.append(token)
        return "/" + "/".join(stack)

C++:

class Solution {
public:
    string simplifyPath(string path) {
        vector<string> v;
        int i = 0;
        while (i < path.size()) {
            while (path[i] == '/' && i < path.size()) ++i;
            if (i == path.size()) break;
            int start = i;
            while (path[i] != '/' && i < path.size()) ++i;
            int end = i - 1;
            string s = path.substr(start, end - start + 1);
            if (s == "..") {
                if (!v.empty()) v.pop_back(); 
            } else if (s != ".") {
                v.push_back(s);
            }
        }
        if (v.empty()) return "/";
        string res;
        for (int i = 0; i < v.size(); ++i) {
            res += '/' + v[i];
        }
        return res;
    }
};