Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:
Input: "aab" Output: [ ["aa","b"], ["a","a","b"] ]
思路问题:自己的思路没问题
结束条件:表示pos到头了
if(pos==s.length()) res.add(new ArrayList<String>(list));
i的for循环里,应该是i + 1
public class Solution { public List<List<String>> partition(String s) { List<List<String>> res = new ArrayList<List<String>>(); List<String> list = new ArrayList<String>(); dfs(s,0,list,res); return res; } public void dfs(String s, int pos, List<String> list, List<List<String>> res){ if(pos==s.length()) res.add(new ArrayList<String>(list));//表示pos到头了 else{ for(int i=pos;i<s.length();i++){ if(isPal(s,pos,i)){ list.add(s.substring(pos,i+1)); dfs(s,i+1,list,res); //i的for循环里,应该是i + 1 list.remove(list.size()-1); } } } } public boolean isPal(String s, int low, int high){ while(low<high) if(s.charAt(low++)!=s.charAt(high--)) return false; return true; } }