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Palindrome Partitioning II
Given a string s, partition s such that every substring of the partition is 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.
class Solution { public: int minCut(string s) { int n=s.length(); int* pali1=new int[n+1]; int* pali2=new int[n+1]; //1 3 5 for(int i=0;i<n;i++) pali1[i]=1; for(int len=3;len<=n;len=len+2) for(int l=len/2;l<=n-1-len/2;l++) if(pali1[l]==len-2 && s[l-len/2]==s[l+len/2]) pali1[l]=len; //0 2 4 for(int i=0;i<n;i++) pali2[i]=0; for(int len=2;len<=n;len=len+2) for(int l=len/2-1;l<=n-1-len/2;l++) if(pali2[l]==len-2 && s[l+1-len/2]==s[l+len/2]) pali2[l]=len; int** cutcache=new int*[n]; for(int i=0;i<n;i++) { cutcache[i]=new int[n]; for(int j=0;j<n;j++) cutcache[i][j]=-1; } return mincut(cutcache,s,0,n-1,pali1,pali2); } int mincut(int** cutcache,const string& s,int l,int r,const int* pali1,const int* pali2) { if(cutcache[l][r]!=-1) return cutcache[l][r]; if(l>=r) return 0; if(check(s,l,r,pali1,pali2)) { cutcache[l][r]=0; return 0; } int cut=r-l+1; for(int i=l;i<r;i++) if(check(s,l,i,pali1,pali2)) { int newcut=1+mincut(cutcache,s,i+1,r,pali1,pali2); if(newcut<cut) cut=newcut; } cutcache[l][r]=cut; return cut; } bool check(const string& s,int l,int r,const int* pali1,const int* pali2) { int len=r-l+1; int mid=(l+r)/2; if(len%2==1 && pali1[mid]>=len) return true; if(len%2==0 && pali2[mid]>=len) return true; return false; } };
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