LeetCode 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:

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

很明显这是一道permutation 题，第一想法就是DFS或者BFS,很多人都直接jump to DFS,但其实BFS也是可以的，可能写起来没有DFS更简洁，但是也不为练习和理解BFS的一道好题

太多人贴DFS的解法了，我只贴一下自己的BFS的解法

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> result = new ArrayList<String>();
        if(n==0) return result;
        Queue<Node> q = new LinkedList<Node>();
        q.offer(new Node(1,0,"("));
        while(q.size()>0){
            Node temp = q.poll();
            if(temp.left <n){
                q.offer(new Node(temp.left+1,temp.right, temp.val +"("));
            }
            if(temp.right<temp.left && temp.right<n){
                 q.offer(new Node(temp.left,temp.right+1, temp.val +")"));
            }
            if(temp.right ==n){
                result.add(temp.val);
            }
        }
        return result;
    }
    
    class Node{
        String val;
        int left;
        int right;
        Node(int left, int right, String val){
            this.left = left;
            this.right =right;
            this.val = val;
        }
    }
}

 

另外这道题还有一种Iterative 的解法

public class Solution {
    public List<String> generateParenthesis(int n) {
        List<List<String>> lists = new ArrayList<>();
        lists.add(Collections.singletonList(""));
        
        for (int i = 1; i <= n; ++i)
        {
            final List<String> list = new ArrayList<>();
            
            for (int j = 0; j < i; ++j)
            {
                for (final String first : lists.get(j))
                {
                    for (final String second : lists.get(i - 1 - j))
                    {
                        list.add("(" + first + ")" + second);
                    }
                }
            }
            
            lists.add(list);
        }
        
        return lists.get(lists.size() - 1);
    }
}