LeetCode#22. Generate Parentheses

题目：

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:

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

题意：

这道题注意一个生成括号很明显的特征，那就是在放置括号的时候，如果剩余的左括号数大于等于剩余的右括号数，这时候一定要放置左括号，不能放置右括号。

一种递归实现的算法如下：

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

输出：vector<string> result

function solution() {

若leftnums = 0 并且 rightnums = 0, 将s加入result中

若leftnums > 0 则递归调用（ s += ‘（’, leftnums -1 ）

若rightnums > 0 则递归调用（ s += ‘）’, rightnums -1）

}

一种c++的代码实现如下：

#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);}}};