Palindrome Partitioning II

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

动态规划 这一类的题目最重要的就是保护现场。。 不能改变现场的情况：

比如这里 就不能

 if(cur < max){

　　 findmin(s,i + 1,cur + 1);

 }
写成：
if(cur < max){
　　cur ++;
　　pointer = i + 1;
　　findmin(s,pointer,cur);
}
会影响后面的同级别的其他操作。

 1 public class Solution {

 2     int max = 0;

 3     public int minCut(String s) {

 4         // Start typing your Java solution below

 5         // DO NOT write main() function

 6         max = s.length() - 1;

 7         int pointer = 0;

 8         if(s.length() == 0){

 9             return 0;

10         }

11         int cur = 0;

12         findmin(s,pointer,cur);

13         int k = max;

14         return max;

15     }

16     public void findmin(String s,int pointer,int cur){

17         int len = s.length();

18         if(pointer == len){

19             if(cur - 1 < max){

20                 max = cur - 1;

21             }

22         }

23         for(int i = pointer; i < len; i ++){

24             int flag = 0;

25             if(i == pointer){

26                 findmin(s,i + 1,cur + 1);

27             }

28             else if((i - pointer + 1) % 2 == 0){

29                 int mid = (i - pointer - 1)/2;

30                 int j = 0;

31                 for(; j < mid + 1; j ++){

32                     if(s.charAt(j+pointer) != s.charAt(i - j)){

33                         flag = 1;

34                         break;

35                     }

36                 }

37                 if(flag == 0){

38                     if(cur < max){

39                         findmin(s,i + 1,cur + 1);

40                     }

41                 }

42             }

43             else{

44                 int mid = (i - pointer) / 2 - 1;

45                 int j = 0;

46                 for(; j < mid + 1; j ++){

47                     if(s.charAt(j+pointer) != s.charAt(i - j)){

48                         flag = 1;

49                         break;

50                     }

51                 }

52                 if(flag == 0){

53                     if(cur < max){

54                         findmin(s,i + 1,cur + 1);

55                     }

56                 }

57             }

58         }

59     }

60 }

 

 第四遍：

分为两步，第一步找到所有是palindrome的substring。 O(n^2).

第二步是算从0- i,的最小的cut数。1—D dp

dp[i] = 0, if status[0][i] = true(从0到i刚好是个palindrome)

dp[i] = min(dp[k - 1] + 1), if(k 到 i 刚好是palindrome， k 在[1, i]中)

 1 public class Solution {

 2     /*

 3         this is an 1-D DP problem

 4         with an 2-D information 

 5         dp[j] = min(dp[k](i <= k <= j - 1) + 1(if [k, j] is palindrome), 0 (if[i, j] is palindrome));

 6         dp[i] = 0;

 7         i - start point; j - end point.

 8     */

 9     public int minCut(String s) {

10         if(s == null || s.length() == 0) return 0;

11         int len = s.length();

12         int[] dp = new int[len];

13         boolean[][] status = new boolean[len][len];

14         for(int i = 0; i < s.length() * 2; i ++){

15             int left = i / 2;

16             int right = (i + 1) / 2;

17             for(; left > -1 && right < s.length(); left --, right ++){

18                 if(s.charAt(left) == s.charAt(right)){

19                     status[left][right] = true;

20                 }else{

21                     break;

22                 }

23             }

24         }

25         for(int i = 0; i < len; i ++){

26             if(status[0][i] == true) dp[i] = 0;

27             else{

28                 dp[i] = Integer.MAX_VALUE;

29                 for(int j = 1; j <= i; j ++){

30                     if(status[j][i] == true){

31                         dp[i] = Math.min(dp[i], dp[j - 1] + 1);

32                     }

33                 }

34             }

35         }

36         return dp[len - 1];

37     }

38 }