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.
#include <iostream>
#include <string>
#include <vector>
using namespace std;
bool check(string& str,int begin,int end)
{
int i = begin ;
int j = end;
for(;i<=j;i++,j--)
if(str[i] != str[j])
return false;
return true;
}
void helper(string& str,int begin,int end,vector<int>& pos,int& count)
{
int i,j,k;
if(begin > end)
{
int k =0;
for(j=0;j<pos.size();j++)
{
if(pos[j] != -1 && j != pos.size()-1)
{
k++;
pos[j] = -1;
}
/*
cout<<str[j];
if(pos[j] != -1)
{
cout<<",";
pos[j] = -1;
} */
}
if(k < count)
count = k;
// cout<<endl;
}
for(i= begin;i<=end;i++)
{
if(check(str,begin,i))
{
pos[i] = i-begin+1;
helper(str,i+1,end,pos,count);
}
}
}
int PalindromePartition(string& str)
{
if(str.length() == 0)
return 0;
vector<int> pos(str.length(),-1);
int count=str.length()-1;
helper(str,0,str.length()-1,pos,count);
return count;
}
int main()
{
string str("aab");
cout<<PalindromePartition(str);
system("pause");
return 0;
}
Palindrome Partitioning II--LeetCode
原文地址：http://blog.csdn.net/yusiguyuan/article/details/44996299
