【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(0,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+1));
                getPartition(i+1,tmp);
                tmp.pop_back();
            }
        }
    }
   
   
   
    void getPalindrome(vector<vector<bool> > &palindrome)
    {
        int startIndex=0;
        int endIndex=n-1;
       
        for(int i=n-1;i>=0;i--)
        {
            for(int j=i;j<n;j++)
            {
                if(i==j)
                {
                    palindrome[i][j]=true;
                }
                else if(j-i==1)
                {
                    palindrome[i][j]=(s[i]==s[j]);
                }
                else if(j-i>1)
                {
                    palindrome[i][j]=(s[i]==s[j]&&palindrome[i+1][j-1]);
                }
            }
        }
    }
};