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:
"((()))", "(()())", "(())()", "()(())", "()()()"
只要左括号数大于1就可以添加左括号。只要右括号数大于左括号数就可以添加右括号。
1 class Solution { 2 public: 3 vector<string> generateParenthesis(int n) { 4 vector<string> res; 5 6 recursive(n, n, "", res); 7 8 return res; 9 } 10 11 void recursive(int n1, int n2, string str, vector<string> &res) { 12 if (n1 == 0 && n2 == 0) { 13 res.push_back(str); 14 return; 15 } 16 17 if (n1 >= 1) { 18 recursive(n1 - 1, n2, str + "(", res); 19 } 20 21 if (n2 > n1) { 22 recursive(n1, n2 - 1, str + ")", res); 23 } 24 } 25 };
网上查了一下，竟然还和Catalan数有关。
通项公式是： \(\frac{(2n)!}{(n+1)!n!}\)
递推公式是 \(C_0=1\ and\ C_{n+1}=\sum\limits^n_{i=0}{C_{i}C_{n-i}}\)
n个+1和n个-1构成2n项\(a_1,a_2,\ldots,a_n\)，其部分和满足\(a_1+a_2+\ldots+a_k\ge{}0,0\le{}k\le{}2n\)的序列个数等于第n个Catalan数\(C_n\)。
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.
用了一个栈去实现。当前哪果是右括号，那么栈顶必须是对应的左括号才行。栈为空的话，也只能push左括号。
1 class Solution { 2 public: 3 bool isValid(string s) { 4 if (s.empty()) return true; 5 6 stack<char> st; 7 8 for (int i = 0; i < s.length(); ++i) { 9 if (st.empty()) { 10 if (s[i] == ‘(‘ || s[i] == ‘[‘ || s[i] == ‘{‘) st.push(s[i]); 11 else return false; 12 } else if (s[i] == ‘(‘ || s[i] == ‘[‘ || s[i] == ‘{‘) { 13 st.push(s[i]); 14 } else if (s[i] == ‘)‘) { 15 if (st.top() == ‘(‘) st.pop(); 16 else return false; 17 } else if (s[i] == ‘]‘) { 18 if (st.top() == ‘[‘) st.pop(); 19 else return false; 20 } else if (s[i] == ‘}‘) { 21 if (st.top() == ‘{‘) st.pop(); 22 else return false; 23 } else { 24 return false; 25 } 26 } 27 return st.empty(); 28 } 29 };
重构一下：
1 class Solution { 2 public: 3 bool isValid(string s) { 4 if (s.empty()) return true; 5 6 stack<char> st; 7 8 for (int i = 0; i < s.length(); ++i) { 9 if (s[i] == ‘(‘ || s[i] == ‘[‘ || s[i] == ‘{‘) { 10 st.push(s[i]); 11 continue; 12 } if (st.empty()) { 13 return false; 14 } 15 16 if (s[i] == ‘)‘ && st.top() != ‘(‘) return false; 17 if (s[i] == ‘]‘ && st.top() != ‘[‘) return false; 18 if (s[i] == ‘}‘ && st.top() != ‘{‘) return false; 19 st.pop(); 20 } 21 return st.empty(); 22 } 23 };
用map再重构，可以再简洁一些。
1 class Solution { 2 public: 3 bool isValid(string s) { 4 if (s.empty()) return true; 5 6 map<char, char> pars; 7 pars[‘)‘] = ‘(‘; 8 pars[‘]‘] = ‘[‘; 9 pars[‘}‘] = ‘{‘; 10 11 stack<char> st; 12 13 for (int i = 0; i < s.length(); ++i) { 14 if (pars.find(s[i]) == pars.end()) { 15 st.push(s[i]); 16 continue; 17 } if (st.empty()) { 18 return false; 19 } 20 21 if (st.top() != pars[s[i]]) return false; 22 st.pop(); 23 } 24 return st.empty(); 25 } 26 };
Leetcode | Parentheses 相关,布布扣,bubuko.com
原文：http://www.cnblogs.com/linyx/p/3718955.html