1.Description
数字 n 代表生成括号的对数,请你设计一个函数,用于能够生成所有可能的并且 有效的 括号组合。
2.Example
输入:n = 3
输出:["((()))","(()())","(())()","()(())","()()()"]
输入:n = 1
输出:["()"]
3.Solution
有两种写法:
1.先一直添加括号,添加到不合适的时候再退回来。
2.在添加前直接判断是否该添加、该添加哪边的括号。(2比1快的多)
(注意使用StringBuilder递归时,回溯之后需要delete掉append添加的括号,而使用String时则不用删除。因为String的值是不变的,String+‘(’并不改变String的值,所以回溯回来String不变。而append会改变StringBuilder的值,因此需要删除append的括号。)
1.String版
public List<String> generateParenthesis(int n) {
List<String> res = new ArrayList<>();
if (n <= 0) return res;
dfs(n, "", res, 0, 0);
return res;
}
private void dfs(int n, String path, List<String> res, int open, int close) {
if (open > n || close > open) return;
if (path.length() == 2 * n) {
res.add(path);
return;
}
dfs(n, path + "(", res, open + 1, close);
dfs(n, path + ")", res, open, close + 1);
}
StringBuilder版
class Solution {
public List<String> generateParenthesis(int n) {
List<String> ans = new ArrayList<String>();
if(n<=0) {
return ans;
}
dfs(ans, new StringBuilder(), n, 0, 0);
return ans;
}
public void dfs(List<String> ans,StringBuilder str,int n,int left,int right) {
if(left>n||right>left) {
return;
}
if(str.length()==2*n) {
ans.add(str.toString());
return;
}
dfs(ans, str.append("("), n, left+1, right);
str.deleteCharAt(str.length()-1);
dfs(ans, str.append(")"), n, left, right+1);
str.deleteCharAt(str.length()-1);
}
}
2.String版
public List<String> generateParenthesis(int n) {
List<String> res = new ArrayList<>();
if (n <= 0) return res;
dfs(n, "", res, 0, 0);
return res;
}
private void dfs(int n, String path, List<String> res, int open, int close) {
if (path.length() == 2 * n) {
res.add(path);
return;
}
if (open < n) {
dfs(n, path + "(", res, open + 1, close);
}
if (close < open) {
dfs(n, path + ")", res, open, close + 1);
}
}
StringBuilder版
class Solution {
public List<String> generateParenthesis(int n) {
List<String> ans = new ArrayList<String>();
backtrack(ans, new StringBuilder(), 0, 0, n);
return ans;
}
public void backtrack(List<String> ans, StringBuilder str, int open, int close, int n) {
if (str.length() == 2*n) {
ans.add(str.toString());
return;
}
if (open < n) {
backtrack(ans, str.append('('), open + 1, close, n);
str.deleteCharAt(str.length() - 1);
}
if (close < open) {
backtrack(ans, str.append(')'), open, close + 1, n);
str.deleteCharAt(str.length() - 1);
}
}
}