Leet Palindrome Partitioning II

 1 class Solution {

 2 public:

 3     int minCut(string s) {

 4         int len = s.length();

 5         int*  p_dp = new int[len + 1];

 6         char* s_dp = new char[len * len];

 7         int*  mm = new int[len + 1];

 8         mm[0] = 0;

 9         memset(s_dp, -1, len * len);

10         memset(p_dp, 0, (len + 1) * sizeof(int));

11 

12         int ret;

13         for (int i=1; i<=len; i++) {

14             int sub_len = i;

15             int minc = INT_MAX;

16 

17             for (int j=0; j<=i-1; j++, sub_len--) {

18 

19                 char* p_is = &s_dp[j * len + i-1];

20                 char b = 0;

21                 if (sub_len >= 3 && -1 != (b = s_dp[(j+1) * len + i - 2])) {

22                     *p_is = b && (s[j] == s[j + sub_len - 1]);

23                 }

24                 if (*p_is == -1) {

25                     int p = j, q = j + sub_len - 1;

26                     for (; p < q && s[p] == s[q]; p++, q--);

27                     *p_is = (p < q) ? 0 : 1;

28                 }

29                 if (*p_is == 0) continue;

30                 if (p_dp[j] < minc) minc = p_dp[j];

31                 if (minc == mm[j]) break;

32             }

33             p_dp [i] = minc + 1;

34             mm[i] = p_dp[i];

35             for (int k = i-1; k >=0 && mm[k]>mm[k+1]; k--) {

36                 mm[k] = mm[k+1];

37             }

38         }

39         ret = p_dp[len];

40         delete[] mm;

41         delete[] p_dp;

42         delete[] s_dp;

43         return ret - 1;

44     }

45 };