LeetCode 131.Palindrome Partitioning

Question

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"]
]

Explanation

用回溯法找到每种可能的组合的模式，然后对于每个部分字符串去判断是否是回文字符串。算是常规题，略组合了一下，有趣。

Code

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> result = new ArrayList<>();
        if (s == null || s.length() == 0) return result;
        List<String> patition = new ArrayList<>();
        helper(s, 0, patition, result);
        return result;

    }

    public void helper(String s, int index, List<String> partition, List<List<String>> result) {
        if (index == s.length()) {
            result.add(new ArrayList<String>(partition));
        }
        for (int i=index; i<s.length(); i++) {
            String subString = s.substring(index, i+1);
            if (!isPalindrome(subString)) continue;
            partition.add(subString);
            helper(s, i+1, partition, result);
            partition.remove(partition.size()-1);

        }
    }

    //check for palindrome
    public boolean isPalindrome(String s) {
        if (s == null || s.length() == 0) return true;
        int i=0; int j = s.length() -1 ;
        s = s.toLowerCase();
        while (i < j) {
            if (!Character.isLetterOrDigit(s.charAt(i))) {
                i ++;
            } else if (!Character.isLetterOrDigit(s.charAt(j))) {
                j--;
            } else if (s.charAt(i)!=s.charAt(j)) {
                return false;
            } else {
                i++; j--;
            }
        }
        return true;
    }

}