代码的优化过程： 生成括号 Generate Parentheses

我是DFS的初学者，所以比较在意一些形式化的东西。现在我一步步得出那样的代码。



版本0： 学习permutation unique 的解法。

5 vector<vector<int> > permuteUnique(vector<int> &num) {

6 vector<vector<int> >result = vector<vector<int> >();

7 vector<int> tmp = vector<int>(); //intemediate result;

8 vector<bool> visit(num.size()); //state

9 int i, step;

10 if (num.size() == 0)

11 return result;

12 sort(num.begin(), num.end());

13 for(i = 0; i < num.size(); i++)

14 visit[i] = false;

15 dfs(result, tmp, num, visit, step);

16 return result;

17 }

18 void dfs(vector<vector<int> > &result, vector<int> tmp, vector<int> num, vector<bool> visit, int step) {

19 if (tmp.size() == num.size()) {

20 result.push_back(tmp);

21 return ;

22 }

23 int i;

24 for(i = 0; i < num.size(); i ++) {

25 // i-1 was visited last time, not this time, the result was same

26 if (visit[i] == true || (i != 0 && num[i] == num[i-1] && !visit[i-1]))

27 continue;

28 tmp.push_back(num[i]);

29 visit[i] = true;

30 dfs(result, tmp, num, visit, step+1);

31 tmp.pop_back();

32 visit[i] = false;

33 }

34 return;

35 }

这里 是DFS最基本的形式。

版本 一: 生成括号

4 vector<string> generateParenthesis(int n) {

5 vector<string> result;

6 string s = string(n, '(');

7 s.append(n,')');

8 string sol;

9 int left, right;

10 left = 0;

11 right = 0;

12 vector<bool> visit(2*n, false);

13 generate(sol, result, s, left, right, visit, n);

14 return result;

15 }

16 void generate(string sol, vector<string> &result, string s, int &left, int &right, vector<bool> visit, int n) {

17 if (left == n && right == n) {

18 result.push_back(sol);

19 return;

20 }

21 int i;

22 for (i = 0; i < s.size(); i ++) {

23 if (visit[i] == true || (i != 0 && s[i] == s[i-1] && !visit[i-1]))

24 continue;

25 if (left == right && s[i] == ')') {

26 continue;

27 }

28 if (s[i] == '(' && !visit[i]) {

29 visit[i] = true;

30 left = left +1;

31 sol.push_back(s[i]);

32 generate(sol, result, s, left, right, visit, n);

33 sol.pop_back();

34 left = left-1;

35 visit[i]= false;

36 }

37 else if (s[i] == ')' && !visit[i] && left > right) {

38 visit[i] = true;

39 right = right + 1;

40 sol.push_back(s[i]);

41 generate(sol, result, s, left, right, visit, n);

42 sol.pop_back();

43 right = right -1;

44 visit[i]= false;

45 }

46 }

47 }

版本2： 优化，循环中只有 2个元素， 所以没有必要搞个循环，只需要列举。

void getParenthesis(string &s, vector<string> &result, int &left, int &right, int n)

{

if(left==n && right==n)

{

result.push_back(s);

return;

}



if(left < n)

{

s.push_back('(');

left = left +1;

getParenthesis(s, result, left, right, n);

left = left -1;

s.pop_back();

}



if(right < left)

{

s.push_back(')');

right = right +1;

getParenthesis(s, result, left, right, n);

right = right -1;

s.pop_back();

}

}

vector<string> generateParenthesis(int n) {

vector<string> result;

if(n==0) return result;

string s;

int left, right;

left = 0;

right = 0;

getParenthesis(s, result, left, right, n);

return result;

}

版本三： 再次精简

void getParenthesis(string s, vector<string> &result, int left, int right, int n)

{

if(left==n && right==n)

{

result.push_back(s);

return;

}

if(left < n)

{

getParenthesis(s + '(', result, left + 1, right, n);

}

if(right < left)

{

getParenthesis(s + ')', result , left, right + 1, n);

}

}

vector<string> generateParenthesis(int n) {

vector<string> result;

if(n==0) return result;

string s;

int left, right;

left = 0;

right = 0;

getParenthesis(s, result, left, right, n);

return result;

}