[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：

View Code

 1 package Palindrome.Partitioning.I;

 2 

 3 import java.util.ArrayList;

 4 

 5 public class calc {

 6 

 7     

 8 

 9     public boolean palindrome(String c) {

10         if(c.length() == 1){

11             return true;

12         }

13         for (int i = 0; i < c.length() / 2; i++) {

14             if (c.charAt(i) != c.charAt(c.length() - 1 - i)) {

15                 return false;

16             }

17         }

18         return true;

19     }

20 

21     public ArrayList<ArrayList<String>> partition(String s) {

22         // Start typing your Java solution below

23         // DO NOT write main() function

24         int sn = s.length();

25         if (sn == 1) {

26             ArrayList<ArrayList<String>> t = new ArrayList<ArrayList<String>>();

27             ArrayList<String> ti = new ArrayList<String>();

28             ti.add(s);

29             t.add(ti);

30             return t;

31         }

32         ArrayList<ArrayList<String>> ret = new ArrayList<ArrayList<String>>();

33         for (int i = 1; i <= s.length(); i++) {

34             String ss = s.substring(0, i);

35 

36             if (palindrome(ss)) {

37                 if (i == s.length()) {

38                     ArrayList<ArrayList<String>> t = new ArrayList<ArrayList<String>>();

39                     ArrayList<String> ti = new ArrayList<String>();

40                     ti.add(s);

41                     t.add(ti);

42                     ret.addAll(t);

43                 }

44                 ArrayList<ArrayList<String>> t = partition(s.substring(i));

45                 // add ss to head of each array-list

46                 for (ArrayList<String> ti : t) {

47                     ti.add(0, ss);

48                 }

49                 ret.addAll(t);

50 

51             }

52         }

53         return ret;

54     }

55 

56     public static void main(String args[]) {

57         calc c = new calc();

58         ArrayList<ArrayList<String>> t = c.partition("aab");

59         for(ArrayList<String> ti : t){

60             for(String tii : ti){

61                 System.out.print(tii + ", ");

62             }

63             System.out.println();

64         }

65     }

66 

67 }