小白笔记----------------------------------leetcode(22. Generate Parentheses )
2017-08-03 11:09
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Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
这道算法涉及到什么卡特兰数,我们把他考虑成一个树的剪枝问题,确定剪枝的条件,就是剩余的左括号树一定大于右括号数。
本文算法,采纳自这篇大牛博客,详细解释,请跳转http://blog.csdn.net/yutianzuijin/article/details/13161721
public class Solution {
public List<String> generateParenthesis(int n) {
String result = "";
List<String> result1 = new LinkedList();//两个规则,一个是要对称,一个是全部情况
result1 = backtrack(n,n,result,result1);
return result1;
}
public List<String> backtrack(int leftnum,int rightnum,String s,List<String> result){
if(leftnum == 0 && rightnum == 0){
result.add(s);
}
if(leftnum > 0){
backtrack(leftnum-1,rightnum,s.concat("("),result);
}
if(rightnum > 0 && leftnum < rightnum){
backtrack(leftnum,rightnum-1,s.concat(")"),result);
}
return result;
}
}
这道算法涉及到什么卡特兰数,我们把他考虑成一个树的剪枝问题,确定剪枝的条件,就是剩余的左括号树一定大于右括号数。
本文算法,采纳自这篇大牛博客,详细解释,请跳转http://blog.csdn.net/yutianzuijin/article/details/13161721
public class Solution {
public List<String> generateParenthesis(int n) {
String result = "";
List<String> result1 = new LinkedList();//两个规则,一个是要对称,一个是全部情况
result1 = backtrack(n,n,result,result1);
return result1;
}
public List<String> backtrack(int leftnum,int rightnum,String s,List<String> result){
if(leftnum == 0 && rightnum == 0){
result.add(s);
}
if(leftnum > 0){
backtrack(leftnum-1,rightnum,s.concat("("),result);
}
if(rightnum > 0 && leftnum < rightnum){
backtrack(leftnum,rightnum-1,s.concat(")"),result);
}
return result;
}
}
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