Generate Parentheses 解答

Question

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:

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

Solution 1

Two conditions to check:

if remained left '(' nums < remained right ')', for next char, we can put '(' or ')'.

if remained left '(' nums = remained right ')', for next char, we can only put '('.

Draw out the solution tree, and do DFS.

 public class Solution {
     public List<String> generateParenthesis(int n) {
         List<String> result = new ArrayList<String>();
         List<Character> list = new ArrayList<Character>();
         dfs(n, n, list, result);
         return result;
     }

     private void dfs(int leftRemain, int rightRemain, List<Character> list, List<String> result) {
         if (leftRemain < 0 || rightRemain < 0 || leftRemain > rightRemain)
             return;

         if (leftRemain == 0 && rightRemain == 0) {
             int size = list.size();
             StringBuilder sb = new StringBuilder(size);
             for (char tmpChar : list)
                 sb.append(tmpChar);
             result.add(sb.toString());
             return;
         }

         if (leftRemain == rightRemain) {
             list.add('(');
             dfs(leftRemain - 1, rightRemain, list, result);
             list.remove(list.size() - 1);
         } else {
             list.add(')');
             dfs(leftRemain, rightRemain - 1, list, result);
             list.remove(list.size() - 1);
             list.add('(');
             dfs(leftRemain - 1, rightRemain, list, result);
             list.remove(list.size() - 1);
         }
     }
 }

Solution 2

A much simpler solution.

 public class Solution {
     public List<String> generateParenthesis(int n) {
         List<String> result = new ArrayList<String>();
         List<Character> list = new ArrayList<Character>();
         dfs(n, n, "", result);
         return result;
     }

     private void dfs(int leftRemain, int rightRemain, String prefix, List<String> result) {
         if (leftRemain < 0 || rightRemain < 0 || leftRemain > rightRemain)
             return;
         if (leftRemain == 0 && rightRemain == 0) {
             result.add(new String(prefix));
             return;
         }
         if (leftRemain > 0)
             dfs(leftRemain - 1, rightRemain, prefix + "(", result);
         if (rightRemain > 0)
             dfs(leftRemain, rightRemain - 1, prefix + ")", result);
     }
 }