20. Valid Parentheses \ 71. Simplify Path

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;    }};

在这里需要注意的是我们使用了stringstream和getline的用法来切割字符串。