【leetcode】Palindrome Partitioning

Palindrome Partitioning

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

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

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

为了加速运算，可以利用动态规划，求出满足回文的子串的位置

 

palindrome[i][j]表示了第字符串中，s[i,i+1,……, j]是否是回文

 

可以有以下递推公式：

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]

 

 

得到了该回文表后，我们利用回溯法得到所有的子串

 

 

 

 class Solution {
 public:

     vector<vector <string> > res;

     vector<vector<bool> > palindrome;
     string s;
     int n;

     vector<vector<string>> partition(string s) {

          this->s=s;
          this->n=s.length();

          vector<vector<bool> > palindrome(n,vector<bool>(n));
          getPalindrome(palindrome);
          this->palindrome=palindrome;

          vector <string> tmp;
          getPartition(,tmp);

          return res;
     }

     //回溯得到子串
     void getPartition(int start,vector<string> tmp)
     {

         if(start==n)
         {
             res.push_back(tmp);
             return;
         }

         for(int i=start;i<n;i++)
         {
             if(palindrome[start][i])
             {
                 tmp.push_back(s.substr(start,i-start+));
                 getPartition(i+,tmp);
                 tmp.pop_back();
             }
         }
     }

     void getPalindrome(vector<vector<bool> > &palindrome)
     {
         int startIndex=;
         int endIndex=n-;

         for(int i=n-;i>=;i--)
         {
             for(int j=i;j<n;j++)
             {
                 if(i==j)
                 {
                     palindrome[i][j]=true;
                 }
                 else if(j-i==)
                 {
                     palindrome[i][j]=(s[i]==s[j]);
                 }
                 else if(j-i>)
                 {
                     palindrome[i][j]=(s[i]==s[j]&&palindrome[i+][j-]);
                 }
             }
         }
     }
 };