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:
"((()))", "(()())", "(())()", "()(())", "()()()"
[解题思路]
典型的递归。一步步构造字符串。当左括号出现次数<n时,就可以放置新的左括号。当右括号出现次数小于左括号出现次数时,就可以放置新的右括号。
[Code]
1: void CombinationPar(vector<string>& result, string& sample, int deep,
2: int n, int leftNum, int rightNum)
3: {
4: if(deep == 2*n)
5: {
6: result.push_back(sample);
7: return;
8: }
9: if(leftNum<n)
10: {
11: sample.push_back('(');
12: CombinationPar(result, sample, deep+1, n, leftNum+1, rightNum);
13: sample.resize(sample.size()-1);
14: }
15: if(rightNum<leftNum)
16: {
17: sample.push_back(')');
18: CombinationPar(result, sample, deep+1, n, leftNum, rightNum+1);
19: sample.resize(sample.size()-1);
20: }
21: }
22: vector<string> generateParenthesis(int n) {
23: // Start typing your C/C++ solution below
24: // DO NOT write int main() function
25: vector<string> result;
26: string sample;
27: if(n!= 0)
28: CombinationPar(result, sample, 0, n, 0, 0);
29: return result;
30: }
这题比较简单,没什么可说的。
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