"Leetcode" Palindrome partitioning II

Palindrome Partitioning II

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.

The method of dynamic programming can first get isp[i][j], that is, whether the string, from I to J is a palindrome isp[i][j]= ((s[i]==s[j)) && (j-i<=1| | ISP[I+1][J-1])) and we then determine the number of splits dp[i] represents the minimum number of splits from I to N, only a point from the middle, so that the minimum number of splits can bedp[i]=min (dp[k]+1) K=I+1......N

1 classSolution {2  Public:3     intMincut (strings) {4        5         intn=s.length ();6vector<vector<BOOL> > IsP (n,vector<BOOL> (n,false));7        8          for(inti=n-1; i>=0; i--)9         {Ten              for(intj=i;j<n;j++) One             { A                 if(s[i]==s[j]&& (j-i<=1|| isp[i+1][j-1])) isp[i][j]=true; -             } -         } the         -vector<int> dp (n+1); -dp[n]=-1;//note the boundary conditions, indicating the dp[i] without splitting, dp[i]=dp[n]+1=0; -          for(inti=n-1; i>=0; i--) +         { -             intmincut=Int_max; +              for(intk=i+1; k<n+1; k++) A             { at                 if(isp[i][k-1]&&mincut>dp[k]+1) -                 { -dp[i]=dp[k]+1; -mincut=Dp[i]; -                 } -             } in         } -         to         returndp[0]; +     } -};

"Leetcode" Palindrome partitioning II