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.

 

这个两个Dp， 一个是对那一部分是palindrome的二维dp。 还有一个是 对cut个数的一维dp。

 1 public class Solution {

 2     int[][] map = null;

 3     public int minCut(String s) {

 4         // IMPORTANT: Please reset any member data you declared, as

 5         // the same Solution instance will be reused for each test case.

 6         map = new int[s.length()][s.length()];

 7         int[] cut = new int[s.length()];

 8         for(int i = 0; i < s.length(); i++)

 9             map[i][i] = 1;

10         for(int i = 0; i < s.length() - 1; i ++){

11             if(s.charAt(i) == s.charAt(i + 1)) map[i][i + 1] = 1;

12             else map[i][i + 1] = -1;

13         }

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

15             cut[i] = i;

16             for(int j = i; j < s.length(); j ++){

17                 map[i][j] = checkPartition(s, i, j);

18             }

19         }

20         for(int j = 0; j < s.length(); j ++){

21             for(int i = 0; i < j; i ++){

22                 if(map[i][j] == 1) cut[j] = Math.min(cut[j], 1 + cut[i]);

23             }

24         }

25         return cut[s.length() - 1];

26     }

27     public int checkPartition(String s, int start, int end){

28         if(map[start][end] != 0) return map[start][end];

29         if(s.charAt(start) != s.charAt(end)) return -1;

30         return checkPartition(s, start + 1, end - 1);

31     }

32 }

 

 1 public class Solution {

 2     public int minCut(String s) {

 3         int leng = s.length();

 4         if(leng == 0 || leng == 1) return 0;

 5         boolean[][] isPal = new boolean[leng][leng];

 6 

 7         int[] dp = new int[leng]; 

 8         for (int i = 0; i < leng; i++) {

 9             dp[i] = leng - 1 - i;

10         }

11 

12         for (int i = leng - 1; i >= 0; --i) {

13             for (int j = i; j < leng; ++j) {

14                 if (s.charAt(i) == s.charAt(j) && (j <= i + 2 || (i + 1 < leng && j - 1 >= 0 && isPal[i + 1][j - 1]))) {

15                     isPal[i][j] = true;

16                     if(j+1 < leng){

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

18                     }else {

19                         dp[i] = 0;

20                     }

21                 }

22             }

23         }

24 

25         return dp[0];

26     }

27 }