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: [ "((()))", "(()())", "(())()", "()(())", "()()()"] Subscribe to see which companies asked this question.
给定 n 对括号,请写一个函数以将其生成新的括号组合,并返回所有组合结果。 样例 给定 n = 3, 可生成的组合如下: "((()))", "(()())", "(())()", "()(())", "()()()"
针对一个长度为2n的合法排列,第1到2n个位置都满足如下规则:左括号的个数大于等于右括号的个数。所以,我们就可以按照这个规则去打印括号:假设在位置k我们还剩余left个左括号和right个右括号,如果left>0,则我们可以直接打印左括号,而不违背规则。能否打印右括号,我们还必须验证left和right的值是否满足规则,如果left>=right,则我们不能打印右括号,因为打印会违背合法排列的规则,否则可以打印右括号。如果left和right均为零,则说明我们已经完成一个合法排列,可以将其打印出来。通过深搜,我们可以很快地解决问题,针对n=2,问题的解空间如下
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public class Solution {
/**
* @param n n pairs
* @return All combinations of well-formed parentheses
*/
public ArrayList<String> generateParenthesis(int n) {
ArrayList<String> result = new ArrayList<String>();
if (n <= 0) {
return result;
}
helper(result, "", n, n);
return result;
}
public void helper(ArrayList<String> result,
String paren, // current paren
int left, // how many left paren we need to add
int right) { // how many right paren we need to add
if (left == 0 && right == 0) {
result.add(paren);
return;
}
if (left > 0) {
helper(result, paren + "(", left - 1, right);
}
if (right > 0 && left < right) {
helper(result, paren + ")", left, right - 1);
}
}
}