Palindrome Partitioning

  [
    ["aa","b"],
    ["a","a","b"]
  ]

 1 class Solution {
 2 public:
 3     vector<vector<string>> partition( string s ) {
 4         if( s.empty() ) { return palindromes; }
 5         size = s.length(); str = s;
 6         vector<bool> *isPalindrome = new vector<bool>[size];
 7         isPalindrome[0].resize( size, true );
 8         for( size_t step = 1; step != size; ++step ) {
 9             isPalindrome[step].resize( size-step, false );
10             if( step == 1 ) {
11                 for( size_t i = 0; i != size-1; ++i ) {
12                     if( s[i] == s[i+1] ) { isPalindrome[1][i] = true; }
13                 }
14             } else {
15                 for( size_t i = 0; i != size-step; ++i ) {
16                     if( s[i] == s[i+step] && isPalindrome[step-2][i+1] ) { isPalindrome[step][i] = true; }
17                 }
18             }
19         }
20         vector<string> palindrome;
21         GeneratePalindromes( palindrome, 0, isPalindrome );
22         delete [] isPalindrome;
23         return palindromes;
24     }
25 private:
26     void GeneratePalindromes( vector<string>& palindrome, int index, vector<bool>* isPalindrome ) {
27         if( index >= size ) { palindromes.push_back( palindrome ); return; }
28         for( int step = 0; step < size-index; ++step ) {
29             if( isPalindrome[step][index] ) {
30                 string subStr = str.substr( index, step+1 );
31                 palindrome.push_back( subStr );
32                 GeneratePalindromes( palindrome, index+step+1, isPalindrome );
33                 palindrome.pop_back();
34             }
35         }
36         return;
37     }
38     int size;
39     string str;
40     vector<vector<string>> palindromes;
41 };