LeetCode - Generate Parentheses
6586 단어 LeetCode
2013.12.7 00:49
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:
"((()))", "(()())", "(())()", "()(())", "()()()"
Solution:
If you're familiar with the Catalan Numbers , you'll see it's a typical solution for this problem.
Since all the combinations are required, we'll use DFS to go through every possible sequence.
Let's consider the recursion in the following way:
1. For a sequence of length 2 * n, there're n '(' and n ')'
2. For a valid sequence, number of '(' will never be less than the number of ')' at any position. If there're more ')' than '(', there must be a mismatch.
3. Always do the recursion from left to right.
Time complexity is H(n) * O(n) = C(2 * n, n)/(n + 1) * n, that's roughly O(n!). Space complexity is not sure yet...
Accepted code:
1 // 1AC, simple recursion will do, but keep your mind clear or it's easy to get things complicated
2 class Solution {
3 public:
4 vector<string> generateParenthesis(int n) {
5 // IMPORTANT: Please reset any member data you declared, as
6 // the same Solution instance will be reused for each test case.
7 result.clear();
8
9 dfs(n, n, "");
10
11 return result;
12 }
13 private:
14 vector<string> result;
15
16 void dfs(int cl, int cr, string pat) {
17 if(cl == 0 && cr == 0){
18 result.push_back(pat);
19 }
20
21 int i;
22 if(cl < cr){
23 if(cl > 0){
24 dfs(cl - 1, cr, pat + "(");
25 }
26 if(cr > 0){
27 dfs(cl, cr - 1, pat + ")");
28 }
29 }else if(cl == cr){
30 if(cl > 0){
31 dfs(cl - 1, cr, pat + "(");
32 }
33 }else{
34 return;
35 }
36 }
37 };