2015.04.01 Leetcode Generate Parentheses

最新推荐文章于 2020-10-09 13:50:19 发布

原创最新推荐文章于 2020-10-09 13:50:19 发布 · 650 阅读

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#algorithm #java #leetcode #backtracking

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本文介绍了一种使用深度优先搜索（DFS）与回溯法来生成所有有效括号组合的方法。针对给定数量的括号对，通过递归地添加左括号和右括号，并确保任何时候左括号的数量都不超过右括号，从而避免无效的括号组合。

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