72. Generate Parentheses && Valid Parentheses

Generate Parentheses

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。

思路：可用于卡特兰数一类题目。

void getParenthesis(vector<string> &vec, string s, int left, int right) {
    if(!right && !left) { vec.push_back(s); return; }
    if(left > 0)
        getParenthesis(vec, s+"(", left-1, right);
    if(right > 0 && left < right)
        getParenthesis(vec, s+")", left, right-1);
}

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> vec;
        getParenthesis(vec, string(), n, n);
        return vec;
    }
};

Valid Parentheses

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.

思路： 栈。对 S 的每个字符检查栈尾，若成对，则出栈，否则，入栈。

class Solution {
public:
    bool isValid(string s) {
        bool ans = true;
        char ch[6] = {‘(‘, ‘{‘, ‘[‘, ‘]‘, ‘}‘, ‘)‘};
        int hash[256] = {0};
        for(int i = 0; i < 6; ++i) hash[ch[i]] = i;
        string s2;
        for(size_t i = 0; i < s.size(); ++i) {
            if(s2 != "" && hash[s2.back()] + hash[s[i]] == 5) s2.pop_back();
            else s2.push_back(s[i]);
        }
        if(s2 != "") ans = false;
        return ans;
    }
};