20. Valid Parentheses \ 71. Simplify Path-优快云博客

本文介绍两种常见算法：括号匹配验证与Unix文件路径简化。括号匹配验证算法通过栈结构检查括号是否正确闭合；路径简化算法则利用栈特性去除无效路径组件，得到规范化的绝对路径。

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Valid Parentheses
- 题目描述
- 代码实现
Simplify Path
- 题目描述
- 代码实现

20. Valid Parentheses

题目描述

括号匹配：

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

代码实现

class Solution {
public:
    bool isValid(string s) {
        int s_len = s.size();
        stack<char> tmp;

        for(int i = 0; i < s_len; i++) {
            if(s[i] == '(' || s[i] == '{' || s[i] == '[') {
                tmp.push(s[i]);        
            }
            else {
                if(tmp.empty()) return false;

                char t = tmp.top();
                tmp.pop();

                if(s[i] == ')' && t != '(') return false;
                if(s[i] == ']' && t != '[') return false;
                if(s[i] == '}' && t != '{') return false;
            }
        }

        return tmp.empty()?true:false;
    }
};

class Solution {
public:
    bool isValid(string s) {
        stack<char> paren;
        for (char& c : s) {
            switch (c) {
                case '(': 
                case '{': 
                case '[': paren.push(c); break;
                case ')': if (paren.empty() || paren.top()!='(') return false; else paren.pop(); break;
                case '}': if (paren.empty() || paren.top()!='{') return false; else paren.pop(); break;
                case ']': if (paren.empty() || paren.top()!='[') return false; else paren.pop(); break;
                default: ; // pass
            }
        }
        return paren.empty() ;
    }
};

71. Simplify Path

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".

代码实现

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

class Solution {
public:
    string simplifyPath(string path) {
        string result="", token;
        stringstream ss(path);
        vector<string> tokens;
        while(getline(ss, token, '/')){
            if(token=="." || token=="") continue;
            else if(token==".."){
                if(tokens.size())  tokens.pop_back();
            }
            else tokens.push_back(token);
        }
        if(!tokens.size()) return "/";
        for(int i=0; i<tokens.size(); ++i)
            result += '/' + tokens[i];
        return result;
    }
};