Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
For example, given s = "aab"
,
Return
[ ["aa","b"], ["a","a","b"]
public class Solution { public List<List<String>> partition(String s) { List<List<String>> result=new LinkedList<List<String>>(); LinkedList<String> output=new LinkedList<String>(); dfs(s,0,output,result); return result; } public void dfs(String s,int start,LinkedList<String> output,List<List<String>> result){ if(start==s.length()){ List<String> list=new LinkedList<String>(); list.addAll(output); result.add(list); return; } for(int i=start;i<s.length();i++){ if(isPalindrome(s,start,i)){ output.add(s.substring(start,i+1)); dfs(s,i+1,output,result); output.pollLast(); } } } public boolean isPalindrome(String s,int i,int j){ while(i<=j){ if(s.charAt(i)==s.charAt(j)){ i++; j--; } else return false; } return true; } }
Palindrome Partitioning,布布扣,bubuko.com
原文:http://blog.csdn.net/dutsoft/article/details/38559529