LeetCode第21题之Generate Parentheses（两种解法）

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C++代码：
解法一（在LeetCode上运行效率高于解法二）：

#include <vector>#include <iostream>#include <string>using namespace std;class Solution {private:    vector<string> res;public:     //leftRemain保存还可以放左括号的数目，rightRemain保存还可以放右括号的数目    void dfs(string state, int leftRemain, int rightRemain)    {        //这种情况括号不匹配        if (leftRemain > rightRemain)        {            return;        }        if (0 == leftRemain && 0 == rightRemain)        {            res.push_back(state);            return;        }        if (leftRemain > 0)        {            dfs(state + "(", leftRemain -1, rightRemain);        }        if (rightRemain >0 )        {            dfs(state + ")", leftRemain, rightRemain-1);        }    }    vector<string> generateParenthesis(int n) {        dfs("",n,n);        return res;    }};int main(){    Solution s;    vector<string> r = s.generateParenthesis(3);    for (vector<string>::iterator it = r.begin();it != r.end();++it)    {        cout<<*it<<endl;    }    return 0;}

解法二：

#include <vector>#include <iostream>#include <string>using namespace std;class Solution {private:    //保存结果    vector<string> res;public:    void fun(int deep, int n, int leftNum, int leftTotalNum, string s)    {        //如果左括号的总数大于n,则该字符串不可能满足要求        if (leftTotalNum  > n)        {            return;        }        //如果到达最底层,则s一定满足题意。因为运行到这里时，leftTotalNum<=n,而leftNum>=0        if (n*2 == deep)        {            res.push_back(s);            return;        }        for (int i=0;i<2;++i)        {            if (0 == i)            {                //在deep+1的位置放左括号                fun(deep+1, n, leftNum+1, leftTotalNum+1, s + "(");            }            else            {                //如果有剩余未匹配的左括号数，才能放右括号                if (leftNum > 0)                {                    fun(deep+1, n, leftNum-1, leftTotalNum, s + ")");                }            }        }    }    vector<string> generateParenthesis(int n) {        //剩余未匹配的左括号数        int leftNum = 0;        //字符串中左括号总数        int leftTotalNum =0;        string s = "";        fun(0, n, leftNum, leftTotalNum, s);        return res;    }};int main(){    Solution s;    vector<string> r = s.generateParenthesis(3);    for (vector<string>::iterator it = r.begin();it != r.end();++it)    {        cout<<*it<<endl;    }    return 0;}

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