2015.04.01 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:
"((()))", "(()())", "(())()", "()(())", "()()()"

  题解： 需要用dfs －backtracking. 
 给定的n为括号对，也就是有n个左括号和n个右括号的组合。
按照顺序尝试，直到左右括号都尝试完了，这就可以得到一个解。

注意： 左括号的数目不能大于右括号。避免“)(” 这样不合法的解出现。

/** *  Generate Parantheses * *  https://leetcode.com/problems/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: *  * "((()))", "(()())", "(())()", "()(())", "()()()" *  * Tags: Backtracking. String */  import java.util.*; public class GenerateParenthesis {    public static void main(String[] args) {        System.out.println(generateParenthesis(3));System.out.println();System.out.println(generateParenthesis(4));    }    /**     * Backtracking     * Helper function use left and right to represent available parentheses     * Initialize left as n, right as n     */    public static List<String> generateParenthesis(int n) {        List<String> res = new ArrayList<String>();String current = new String();  // current result        if (n <= 0) return res;        dfs(res, current, n, n);        return res;    }    /**     * @param left available left parentheses     * @param right available right parentheses     * @param current   current result     * @param res   the result list of the problem     */    public static void dfs( List<String> res, String current, int left, int right) {if (left > right) // deal with ")("return;         if (left == 0 && right == 0) {            res.add( new String(current));            return;        }        if (left > 0) dfs( res, current + "(", left-1, right); // add (, right + 1        if (right > 0) dfs(res , current+ ")", left, right-1); // add ), right - 1    }}

参考

http://www.cnblogs.com/springfor/p/3886559.html