【JAVA、C++】LeetCode 022 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:

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

解题思路一：

通过观察n=2和n=3的情况可以知道，只要在n=2的String 开头、末尾、'(' 插入“（）”即可，注意防止重复。

JAVA实现如下：

static public List<String> generateParenthesis(int n) {
		HashSet<String> set = new HashSet<String>();
		if (n == 1)
			set.add("()");
		if (n > 1) {
			ArrayList<String> lastList = new ArrayList<String>(
					generateParenthesis(n - 1));
			StringBuilder sb = new StringBuilder();
			for (String string : lastList) {
				sb = new StringBuilder(string);
				set.add(sb.toString() + "()");
				set.add("()" + sb.toString());
				for (int i = 0; i < sb.length() - 1; i++) {
					if (sb.charAt(i) == '(') {
						sb.insert(i + 1, "()");
						set.add(sb.toString());
					}
					sb = new StringBuilder(string);
				}
			}
		}
		return new ArrayList<String>(set);
	}

解题思路二：

通过观察可以发现，任何符合条件的Parenthesis(n)总是可以分解为(generateParenthesis(k))+generateParenthesis(n-1-k)，同时由于后半部分generateParenthesis(n-1-k)的唯一性，这种算法不会产生重复的元素，因此采用DFS进行一次遍历即可，这种算法更高效！JAVA实现如下：

static public List<String> generateParenthesis(int n) {
		List<String> list = new ArrayList<String>(), leftList, rightList;
		if (n == 0)
			list.add("");// 很关键，不能删除
		if (n == 1)
			list.add("()");
		else {
			for (int i = 0; i < n; i++) {
				leftList = generateParenthesis(i);
				rightList = generateParenthesis(n - i - 1);
				for (String leftPart:leftList)
					for (String rightPart:rightList)
						list.add("(" + leftPart + ")" + rightPart);
			}
		}
		return list;
	}

C++：

  class Solution {
  public:
      vector<string> generateParenthesis(int n) {
          vector<string>res;
          if (n == ) {
              res.push_back("");
              return res;
          }
          if (n == ) {
              res.push_back("()");
              return res;
          }
          for (int i = ; i < n; i++) {
              vector<string> left = generateParenthesis(i);
              vector<string>right = generateParenthesis(n-i-);
              for (string leftPart : left)
                  for (string rightPart : right)
                      res.push_back("(" + leftPart + ")" + rightPart);
          }
          return res;
      }
  };