LeetCode 71. Simplify Path

最新推荐文章于 2025-09-10 22:03:14 发布

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最新推荐文章于 2025-09-10 22:03:14 发布

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分类专栏：算法与OJ C/C++ 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/zhyh1435589631/article/details/51222874

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本文介绍了一种简化Unix风格文件路径的方法，通过分析问题并利用栈结构处理路径中的特殊符号，最终输出规范化的绝对路径。

1. 题目描述

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

For example,
path = “/home/”, => “/home”
path = “/a/./b/../../c/”, => “/c”
click to show corner cases.

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”.

2. 解题思路

这是一道字符串的处理问题，拿到这个问题，首先想到的使用栈结构进行处理，而且通过分析，我们知道，我们所需要处理的内容主要分为 4 个部分：
/ , /. , 这两个部分可以直接忽略
/.. , 弹栈
/xx，压栈
最后全部输出即可

3. code

class Solution {
public:
    string simplifyPath(string path) {
        stack<string> mystack;
        int pos = 0, last = 0;
        string substr;
        while ((pos = path.find('/', last)) < (int)path.size()){
            // 提取字串
            if (pos == -1)
                substr = path.substr(last);
            else
                substr = path.substr(last, pos - last);

            // 处理 /. /.. / /xx
            if (substr == "" || substr == ".")
                ;
            else if (substr == ".."){
                if (!mystack.empty())
                    mystack.pop();
            }
            else{
                mystack.push(substr);
            }

            if (pos == -1)
                break;

            last = pos + 1;
        }

        if (mystack.empty())
            return "/";

        string res;
        while (!mystack.empty()){
            res = "/" + mystack.top() + res;
            mystack.pop();
        }
        return res;
    }
};

4. 大神解法

一样的思路，不过借助stringstream 实现字符串分隔， so brilliant ！！！

string simplifyPath(string path) {
    string res, tmp;
    vector<string> stk;
    stringstream ss(path);
    while(getline(ss,tmp,'/')) {
        if (tmp == "" or tmp == ".") continue;
        if (tmp == ".." and !stk.empty()) stk.pop_back();
        else if (tmp != "..") stk.push_back(tmp);
    }
    for(auto str : stk) res += "/"+str;
    return res.empty() ? "/" : res;
}