## LeetCode

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

Return

*s*=`"aab"`

,Return

[ ["aa","b"], ["a","a","b"] ]

Solution: DFS, Recursion

declare start point st, make it move from 0-> s.length(), for each st, check every substring generated by st and i, st+1<=i<=s.length(), if the substring is a palindrome then put it into ArrayList<String> partition, if st reach s.length() mean we got a valid partition for given string s, then put it into ArrayList<ArrayList<String>> result;