"Leetcode" palindrome partitioning

Palindrome Partitioning

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

Return all possible palindrome partitioning of s.

For example, given s = "aab" ,
Return

  [    ["AA", "B"],    ["A", "a", "B"]  ]

In order to accelerate the operation, the dynamic programming can be used to find the position of the substring satisfying the palindrome palindrome[i][j] represents the string, S[i,i+1,......, j] is a palindrome can have the following recursive formula: if (I==J) palindrome[i][j]=true  ; if (i-j=1) palindrome[i][j]=s[i]==s[j];if (i-j>1) palindrome[i][j]=palindrome[i+1][j-1]&&s[i]==s[j] After getting the palindrome table, we use backtracking to get all the substrings

1 classSolution {2  Public:3  4Vector<vector <string> >Res;5    6vector<vector<BOOL> >palindrome;7     strings;8     intN;9    Tenvector<vector<string>> partition (strings) { One         A           This->s=s; -           This->n=s.length (); -           thevector<vector<BOOL> > Palindrome (n,vector<BOOL>(n)); - Getpalindrome (palindrome); -           This->palindrome=palindrome; -           +Vector <string>tmp; -Getpartition (0, TMP); +           A          returnRes; at     } -     -     //backtracking to get substrings -     voidGetpartition (intstart,vector<string>tmp) -     { -   in         if(start==N) -         { to Res.push_back (TMP); +             return; -         } the         *          for(inti=start;i<n;i++) $         {Panax Notoginseng             if(Palindrome[start][i]) -             { theTmp.push_back (S.substr (start,i-start+1)); +Getpartition (i+1, TMP); A Tmp.pop_back (); the             } +         } -     } $     $     -     -     voidGetpalindrome (vector<vector<BOOL> > &palindrome) the     { -         intstartindex=0;Wuyi         intendindex=n-1; the         -          for(inti=n-1; i>=0; i--) Wu         { -              for(intj=i;j<n;j++) About             { $                 if(i==j) -                 { -palindrome[i][j]=true; -                 } A                 Else if(j-i==1) +                 { thePalindrome[i][j]= (s[i]==s[j]); -                 } $                 Else if(j-i>1) the                 { thePalindrome[i][j]= (s[i]==s[j]&&palindrome[i+1][j-1]); the                 } the             } -         } in     } the};

"Leetcode" palindrome partitioning