[leetcode]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"]
  ]

求解所有可能的回文划分。
分析:p=pp(s[1...n])
初始: if n==1, return <<s>>
递归：p=s[1] || pp(s[2...n])
　　　[+s[1..2] || pp(s[3..n]), if s[1..2] is palindrome;
　　　 +s[1..3] || pp(s[4..n]), if s[1..3] is palindrome;
　　　　...
　　　 +s[1..n], if s[1..n] is palindrome;
　　　]
优化：pp(s[i..j])在递归过程中回重复调用，维护Look-up table.
非优化的代码：

package Palindrome.Partitioning.I;

import java.util.ArrayList;

public class calc {

    

    public boolean palindrome(String c) {
        if(c.length() == 1){
            return true;
        }
        for (int i = 0; i < c.length() / 2; i++) {
            if (c.charAt(i) != c.charAt(c.length() - 1 - i)) {
                return false;
            }
        }
        return true;
    }

    public ArrayList<ArrayList<String>> partition(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int sn = s.length();
        if (sn == 1) {
            ArrayList<ArrayList<String>> t = new ArrayList<ArrayList<String>>();
            ArrayList<String> ti = new ArrayList<String>();
            ti.add(s);
            t.add(ti);
            return t;
        }
        ArrayList<ArrayList<String>> ret = new ArrayList<ArrayList<String>>();
        for (int i = 1; i <= s.length(); i++) {
            String ss = s.substring(0, i);

            if (palindrome(ss)) {
                if (i == s.length()) {
                    ArrayList<ArrayList<String>> t = new ArrayList<ArrayList<String>>();
                    ArrayList<String> ti = new ArrayList<String>();
                    ti.add(s);
                    t.add(ti);
                    ret.addAll(t);
                }
                ArrayList<ArrayList<String>> t = partition(s.substring(i));
                // add ss to head of each array-list
                for (ArrayList<String> ti : t) {
                    ti.add(0, ss);
                }
                ret.addAll(t);

            }
        }
        return ret;
    }

    public static void main(String args[]) {
        calc c = new calc();
        ArrayList<ArrayList<String>> t = c.partition("aab");
        for(ArrayList<String> ti : t){
            for(String tii : ti){
                System.out.print(tii + ", ");
            }
            System.out.println();
        }
    }

}