Leetcode Generate Parentheses
2016-06-29 05:15
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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:
Difficulty: Medium
public class Solution {
List<String> res = new ArrayList<String>();
public void helper(int left, int right, String s){
if(right < left) return;
if(left == 0 && right == 0){
res.add(new String(s));
return;
}
if(left == right){
s = s + '(';
helper(left - 1, right, s);
s = s.substring(0, s.length() - 1);
return;
}
if(left == 0){
s = s + ')';
helper(left, right - 1, s);
s = s.substring(0, s.length() - 1);
return;
}
s = s + '(';
helper(left - 1, right, s);
s = s.substring(0, s.length() - 1);
s = s + ')';
helper(left, right - 1, s);
s = s.substring(0, s.length() - 1);
return;
}
public List<String> generateParenthesis(int n) {
String s = "";
helper(n, n, s);
return res;
}
}
For example, given n = 3, a solution set is:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
Difficulty: Medium
public class Solution {
List<String> res = new ArrayList<String>();
public void helper(int left, int right, String s){
if(right < left) return;
if(left == 0 && right == 0){
res.add(new String(s));
return;
}
if(left == right){
s = s + '(';
helper(left - 1, right, s);
s = s.substring(0, s.length() - 1);
return;
}
if(left == 0){
s = s + ')';
helper(left, right - 1, s);
s = s.substring(0, s.length() - 1);
return;
}
s = s + '(';
helper(left - 1, right, s);
s = s.substring(0, s.length() - 1);
s = s + ')';
helper(left, right - 1, s);
s = s.substring(0, s.length() - 1);
return;
}
public List<String> generateParenthesis(int n) {
String s = "";
helper(n, n, s);
return res;
}
}
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