LeetCode第21题之Generate Parentheses（两种解法）

#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;
}

#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;
}