LeetCode#22. Generate Parentheses

題目：

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:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

題意：

這道題注意一個生成括號很明顯的特徵，那就是在放置括號的時候，如果剩餘的左括號數大於等於剩餘的右括號數，這時候一定要放置左括號，不能放置右括號。

一種遞歸實現的算法如下：

輸入：vector<string> result, string s = "", int n, int leftnums = n, int rightnums = n;

輸出：vector<string> result

function solution() {

若leftnums = 0 並且 rightnums = 0, 將s加入result中

若leftnums > 0 則遞歸調用（ s += ‘（’, leftnums -1 ）

若rightnums > 0 則遞歸調用（ s += ‘）’, rightnums -1）

}

一種c++的代碼實現如下：

#include<iostream>
#include<vector>
#include<string>

using namespace std;

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> result;
        string s = "";
        int left_bracket_nums = n;
        int right_bracket_nums = n;
        recursion_Parenthesis(result,s,left_bracket_nums,right_bracket_nums);
        return result;
    }
    void recursion_Parenthesis(vector<string> &result, string s, int left_bracket_nums, int right_bracket_nums) {
    	if(left_bracket_nums == 0 && right_bracket_nums == 0) {
    		result.push_back(s);
		}
		if(left_bracket_nums > 0) {
			recursion_Parenthesis(result,s+'(',left_bracket_nums-1,right_bracket_nums);
		} 
		if(right_bracket_nums > 0 && left_bracket_nums < right_bracket_nums) {
			recursion_Parenthesis(result,s+')',left_bracket_nums,right_bracket_nums-1);
		}
	}
};