回溯法（1）

原题：

/**
 * Created by pradhang on 3/15/2017.
 * 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"]
 * ]
 */

答案：

public class PalindromePartitioning {
    /**
     * Main method
     *
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception {
        List> result = new PalindromePartitioning().partition("aaaaaaaaaaaaaaaaaa");
    }

    public List> partition(String s) {
        List> result = new ArrayList<>();
        doNext(0, new ArrayList<>(), s, result);
        return result;
    }

    private void doNext(int i, List row, String s, List> result) {
        if (i == s.length()) {
            List list = new ArrayList<>(row);
            result.add(list);
        } else {
            for (int j = i, l = s.length(); j < l; j++) {
                String sbStr = s.substring(i, j + 1);
                if (isPalindrome(sbStr)) {
                    row.add(sbStr);
                    doNext(j + 1, row, s, result);
                    row.remove(row.size() - 1);
                }
            }
        }
    }

    private boolean isPalindrome(String s) {
        int i = 0, j = s.length() - 1;
        while (i <= j) {
            if (s.charAt(i) != s.charAt(j))
                return false;
            i++;
            j--;
        }
        return true;
    }
}