[leetcode]Palindrome Partitioning

Palindrome Partitioning

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"]
  ]

算法思路：
dfs

public class Solution {
    List<List<String>> result = new ArrayList<List<String>>();
    
    public List<List<String>> partition(String s) {
        if(s == null || s.length() == 0) return result;
        dfs(new ArrayList<String>(),s);
        return result;
    }
    private void dfs(List<String> list,String s){
        if(s.length() == 0){
            result.add(new ArrayList<String>(list));
            return;
        }
        for(int i = 0; i < s.length(); i++){
            String pre = s.substring(0, i + 1);
            if(isPalindrome(pre)){
                list.add(pre);
                dfs(list, s.substring(i + 1));
                list.remove(list.size() - 1);
            }
        }
    }
    private boolean isPalindrome(String s){
        for(int i = 0; i <= s.length() / 2; i++){
            if(s.charAt(i) != s.charAt(s.length() - 1 - i)) return false;
        }
        return true;
    }
}

jd的做法几乎一样，并在博文下面借鉴了两篇其他博文，其中有个DP算法可以看一下。