"Leetcode" palindrome partitioning II (hard) ☆

Given A string s, partition s such that every substring of the partition are 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.

Cut a few knives at least to make each part of a string a palindrome.

Ideas:

Use Cut[i] to store substrings from s[0] ~ S[i-1], and to cut a few knives in the shortest.

For convenience make cut[0] =-1

No cutter required when only one character cut[1] = 0

In other cases, assuming a cutter from ( -1 ~ i-2), if the second part of the S-cutter is a palindrome, cut[i] = cut[j] + 1; Cut[i] Take all the smallest of the tangent

The code is as follows: This is the O (N3) method, the result is timed out ....

classSolution { Public:     intMincut (strings) {vector<int> Cut (s.length () +1,0); cut[0] = -1;  for(inti =2; I <= s.length (); i++)         {             intMinnum =s.length ();  for(intj =0; J < I; J + +)             {                 if(Ispalindrome (S.substr (J, I-J )) {intnum = Cut[j] +1; Minnum=min (num, minnum); }} Cut[i]=Minnum; }         returncut[s.length ()]; }    BOOLIspalindrome (strings) {inti =0, j = s.length ()-1;  while(I <j) {if(S[i]! = S[j])return false; I++; j--; }        return true; }};

All kinds of interception can not pass, had to look at other people's thinking, the original in the judgment palindrome here to make efforts, I am every time I judge it is not a palindrome, in fact, can be previously asked to save the palindrome information, to facilitate the judgment behind.

O (N2) solution

classSolution { Public:intMincut (strings) {vector<int> Cut (s.length () +1,0); Vector<vector<BOOL>> Ispalindrome (s.length () +1, vector<BOOL> (s.length () +1,false)); cut[0] = -1;  for(inti =2; I <= s.length (); i++)         {             intMinnum =s.length ();  for(intj = i-1; J >=0; j--)             {                 if((s[j] = = s[i-1]) && (i-1-J <2|| Ispalindrome[j +1][i-2]) {ispalindrome[j][i-1] =true; Minnum= Min (Cut[j] +1, Minnum); }} Cut[i]=Minnum; }         returncut[s.length ()]; }};

There are more powerful, the above method saved the judgment palindrome information, there is a need not to save, very fast

classSolution { Public:    intMincut (strings) {intn =s.size (); Vector<int> Cut (n+1,0);//Number of cuts for the first K characters         for(inti =0; I <= N; i++) Cut[i] = i1;  for(inti =0; I < n; i++) {             for(intj =0; I-j >=0&& i+j < n && S[i-j]==s[i+j]; J + +)//Odd Length palindromecut[i+j+1] = min (cut[i+j+1],1+cut[i-J]);  for(intj =1; i-j+1>=0&& i+j < n && s[i-j+1] = = S[i+j]; J + +)//even length palindromecut[i+j+1] = min (cut[i+j+1],1+cut[i-j+1]); }        returnCut[n]; }};

"Leetcode" palindrome partitioning II (hard) ☆