LeetCode 22: Generate Parentheses
2015-10-21 17:12
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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:
"((()))", "(()())", "(())()", "()(())", "()()()"
解题思路
思路一:该问题解的个数就是卡特兰数[1],但是现在不是求个数,而是要将所有合法的括号排列打印出来。对一个长度为2n 的合法排列,第1到2n的位置都满足如下规则:左括号的个数大于等于右括号的个数。因此们可以按照这个规则去打印括号:假设在位置 k 我们还剩余 LP 个左括号和 RP 个右括号,如果LP > 0,则我们可以直接打印左括号;能否打印右括号,我们还必须验证 LP 和 RP 的值是否满足规则,如果LP < RP,则我们可以打印右括号,否则不可以打印右括号。如果 LP 和 RP 均为零,则说明我们已经完成一个合法排列,可以将其打印出来。通过DFS,我们可以很快地解决问题。
代码如下:
class Solution { private: void generate(int LP,int RP,string s,vector<string> &result) { if(LP == 0 && RP == 0) { // LP 和 RP 均为零,则说明我们已经完成一个合法排列 result.push_back(s); } if(LP > 0) { // LP > 0,可以打印左括号 generate(LP - 1, RP, s + '(', result); } if(RP > 0 && LP < RP) { // LP < RP,可以打印右括号 generate(LP, RP - 1, s + ')', result); } } public: vector<string> generateParenthesis(int n) { vector<string> result; generate(n, n, "", result); return result; } };
思路二:
class Solution { public: vector<string> generateParenthesis (int n) { if (n == 0) return vector<string>(1, ""); if (n == 1) return vector<string> (1, "()"); vector<string> result; for (int i = 0; i < n; ++i) for (auto inner : generateParenthesis (i)) for (auto outer : generateParenthesis (n - 1 - i)) result.push_back ("(" + inner + ")" + outer); return result; } };请参考百度百科中卡特兰数词条 ↩
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