面试题37：字符串中的括号

字符串经常出现括号的问题，主要就是产生括号对、判断括号对是否有效、最长的括号对子串。

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:

"((()))", "(()())", "(())()", "()(())", "()()()"

class Solution {
public:
    void helper(vector<string>& res,string path,int left,int right,int n){
        if(right > left || left > n || right >n) return;
        if(left == n){
            while(right < n){
                path.push_back(')');
                right++;
            }
            res.push_back(path);
            return;
        }
        path.push_back('(');
        helper(res,path,left+1,right,n);
        path.pop_back();
        path.push_back(')');
        helper(res,path,left,right+1,n);
    }
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        string path;
        helper(res,path,0,0,n);
        return res;
    }
};

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.

class Solution {
public:
    bool isValid(string s) {
        stack<char> st;
        for(size_t i=0;i<s.length();i++){
            if(s[i] == '(' || s[i] == '{' || s[i] == '['){
                st.push(s[i]);
            }else{
                if(st.empty()) return false;
                if(s[i] == ')' && st.top() == '('){
                    st.pop();
                }else if(s[i] == ']' && st.top() == '['){
                    st.pop();
                }else if(s[i] == '}' & st.top() == '{'){
                    st.pop();
                }else{
                    return false;
                }
            }
        }
        return st.empty();

    }
};

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.

class Solution{
public:
    int longestValidParentheses(string s){
        stack<int> lefts; // index of left parentheses.
        int len = s.length();
        if(len < 2) return 0;

        int start = -1;
        int res = 0;
        for (int i=0;i<len;i++){
            if(s[i] == '('){
                lefts.push(i);
            }else{
                if(lefts.empty()){
                    start = i;
                }else{
                    lefts.pop();
                    if(lefts.empty()){
                        res = max(res,i - start);
                    }else{
                        res = max(res,i- lefts.top());
                    }
                }
            }
        }
        return res;
    }
};