Generate Parentheses

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:

"((()))", "(()())", "(())()", "()(())", "()()()"

public class Solution {
    ArrayList<String> result = null;
    public ArrayList<String> generateParenthesis(int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        StringBuffer st = new StringBuffer();
        result = new ArrayList<String>();
        if(n == 0) return result;
        getRow(st, n , n, 0);
        return result;
    }
    public void getRow(StringBuffer st, int n, int m, int num){
        if(n == 0 && m == 0){
            result.add(st.toString());
        }
        if(n > 0){
            st.append('(');
            num ++;
            getRow(st, n - 1, m, num);
            num --;
            st.deleteCharAt(st.length() - 1);
        }
        if(m > 0 && num > 0){
            st.append(')');
            num --;
            getRow(st, n, m - 1, num);
            num ++;
            st.deleteCharAt(st.length() - 1);
        }
    }
}

第三遍：

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> result = new ArrayList<String>();
        generate(result, 0, n, n, new StringBuilder());
        return result;
    }
    
    public void generate(List<String> result,int cur,int left,int right, StringBuilder sb){
        if(left == 0 && right == 0) result.add(sb.toString());
        if(left > 0){
            StringBuilder sbs = new StringBuilder(sb);
            sbs.append('(');
            generate(result, cur + 1, left - 1, right, sbs);
        }
        if(right > 0 && cur > 0){
            StringBuilder sbs = new StringBuilder(sb);
            sbs.append(')');
            generate(result, cur - 1, left, right - 1, sbs);
        }
    }
}