Generate Parentheses
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\(C_n\)。
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.
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 };