22.Generate-Parentheses

22. Generate Parentheses

题目地址

https://leetcode.com/problems/generate-parentheses/

题目描述

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:

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

代码

Approach 1: Brute Force

class Solution {
  public List<String> generateParenthesis(int n) {
        List<String> combinations = new ArrayList();
    generateAll(new char[2 * n], 0, combinations);
    return combinations;
  }

  private void generateAll(char[] current, int pos, List<String> result) {
    if (pos == current.length) {
        if (valid(current)) {
          result.add(new String(current));
        }
      } else {
      /// looks like backtrace
        current[pos] = '(';
        generateAll(current, pos + 1, result);
        current[pos] = ')';
        generateAll(current, pos + 1, result);
      }
    }

    private boolean valid(char[] current) {
      int balance = 0;
      for (char c : current) {
        if (c == '(') {
          balance++;
        } else {
          balance--;
        }
        if (balance < 0)    return false;
      }

      return (balance == 0);
    }

}

Approach #2 Backtracking

class Solution {
  public List<String> generateParenthesis(int n) {
    List<String> ans = new ArrayList();
    backtrack(ans, "", 0, 0, n);
    return ans;
  }

  public void backtrack(List<String> ans, String cur, int open, int close, int max) {
    if (cur.length() == max * 2) {
      ans.add(cur);
      return;
    }

    if (open < max) {
      backtrack(ans, cur + "(", open + 1, close, max);
    }

    if (close < open) {
      backtrack(ans, cur + ")", open, close + 1, max);
    }

  }

}

Approach #3 Closure Number

class Solution {
  public List<String> generateParenthesis(int n) {
    List<String> ans = new ArrayList();
    if (n == 0) {
      ans.add("");
    } else {
      for (int c = 0; c < n; c++) {
        for (String left: generateParenthesis(c)) {
          for (String right: generateParenthesis(n - 1- c)) {
            ans.add("(" + left + ")" + right);
          }
        }
      }
    }
    return ans;
  }
}