括号匹配问题

1.简单括号的匹配

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.

用数据结构栈即可完成算法

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();
    }
};

2.产生括号匹配

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]

产生括号经典的方法就是回溯

class Solution {
public:
    void backtrack(vector<string>&res,string s,int open,int close,int n)
    {
        if(s.size()==2*n)
        {
            res.push_back(s);
            return;
        }
        if(open<max)
            backtrack(res,s+')',open+1,close,n);
        if(close<open)
            backtrack(res,s+'(',open,close+1,n);
    }
    vector<string> generateParenthesis(int n) {
        vector<string>res;
        backtrack(res,"",n,0);
        return res;
    }
};

3.最长括号匹配

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.

For "(()", the longest valid parentheses substring is "()", which has length = 2.

Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4.

这个方法即是把匹配不上的两个括号储存在stack中，最后再比较两个的大小

注意从-1的位置到第一个未匹配的括号还是要比较的，所以就有了 res=max(res,b); 这一句

class Solution {
public:
    int longestValidParentheses(string s) {
        stack<int>m;
        for(int i = 0;i<s.size();++i)
        {
            if(s[i]=='(')
                m.push(i);
            else
            {
                if(!m.empty()&&s[m.top()]=='(')
                    m.pop();
                else
                    m.push(i);
            }
        }
        if(m.empty())
            return s.size();
        int res=0;
        int a=0,b=s.size();
        while(!m.empty())
        {
            a=m.top();
            m.pop();
            res=max(res,b-a-1);
            b=a;
        }
        res=max(res,b);
        return res=res>b-a-1?res:b-a-1;
    }
};