Generate Parentheses
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:
"((()))", "(()())", "(())()", "()(())", "()()()"
分析
小括号串是一个递归结构,跟单链表、二叉树等递归结构一样,首先想到用递归。
一步步构造字符串。当左括号出现次数<n
时,就可以放置新的左括号。当右括号出现次数小于左括号出现次数时,就可以放置新的右括号。
代码1
// Generate Parentheses
// 时间复杂度O(TODO),空间复杂度O(n)
public class Solution {
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<>();
StringBuilder path = new StringBuilder();
if (n > 0) generate(n, path, result, 0, 0);
return result;
}
// l 表示 ( 出现的次数, r 表示 ) 出现的次数
private static void generate(int n, StringBuilder path,
List<String> result, int l, int r) {
if (l == n) {
StringBuilder sb = new StringBuilder(path);
for (int i = 0; i < n - r; ++i) sb.append(')');
result.add(sb.toString());
return;
}
path.append('(');
generate(n, path, result, l + 1, r);
path.deleteCharAt(path.length() - 1);
if (l > r) {
path.append(')');
generate(n, path, result, l, r + 1);
path.deleteCharAt(path.length() - 1);
}
}
}
代码2
另一种递归写法,更加简洁。
// Generate Parentheses
// @author 连城 (http://weibo.com/lianchengzju)
public class Solution {
public List<String> generateParenthesis(int n) {
if (n == 0) return new ArrayList<>(Arrays.asList(""));
if (n == 1) return new ArrayList<>(Arrays.asList("()"));
List<String> result = new ArrayList<>();
for (int i = 0; i < n; ++i)
for (String inner : generateParenthesis (i))
for (String outer : generateParenthesis (n - 1 - i))
result.add ("(" + inner + ")" + outer);
return result;
}
}
相关题目
原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/dfs/generate-parentheses.html