题目一：
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"]
]
解题思路：
1）用一个二维数组来存取两者之间是否是回文。
2）从字符串尾部开始存入可能的回文串。
代码：
class Solution {
public:
vector<vector<string>> partition(string s)
{
int len = s.size();
vector<vector<bool>>f(len+1,vector<bool>(len));
for (int i=1;i<len+1;i++)
{
for (int j = i-1;j>=0;j--)
{
if(ispalindrome(s,j,i-1)==1)
{
f[i][j]=true;
}
}
}
dealstring(s,len,f);
return m_str;
}
private:
vector<vector<string>>m_str;
vector<string>m_temstr;
void dealstring(string s,int count,const vector<vector<bool>>&f)
{
if (count == 0)
{
vector<string>oneanswer(m_temstr.rbegin(),m_temstr.rend());
m_str.push_back(oneanswer);
}
for (int i=0;i<count;i++)
{
if (f[count][i])
{
m_temstr.push_back(s.substr(i,count-i));
dealstring(s,i,f);
m_temstr.pop_back();
}
}
}
int ispalindrome(string s,int i,int j)
{
while (i<j)
{
if (s[i]!=s[j])
{
return 0;
}
i++;
j--;
}
return 1;
}
};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.
解题思路：
1）用二维数组存入判断两者之间是否是回文。
2）同时用一个一维数组存取从这个点开始往右的最小切数。
代码：
class Solution {
public:
int minCut(string s)
{
int size = s.size();
int mincount[size+1];
vector<vector<bool> >m_bool(size,vector<bool>(size));
for(int i=0;i<=size;i++)
{
mincount[i] = size-1-i;
}
for(int j=size-1;j>=0;j--)
{
for(int k = j;k<size;k++)
if(s[j]==s[k]&&(k-j<2||m_bool[j+1][k-1]))
{
m_bool[j][k]=true;
mincount[j]=min(mincount[j],mincount[k+1]+1);
}
}
return mincount[0];
}
};leetcode题目：Palindrome Partitioning 和Palindrome Partitioning II,布布扣,bubuko.com
leetcode题目：Palindrome Partitioning 和Palindrome Partitioning II
原文地址：http://blog.csdn.net/woailiyin/article/details/25724207
