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
2015.04.01 Leetcode Generate Parentheses
原文:http://blog.csdn.net/hustbill/article/details/44830241