Palindrome Partitioning II

题目：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.

算法思想：典型的动态规划问题（如果不知道动态规划，自行百度一下），

令s(i,j)表示字符串s索引从i 到j 分隔出来的子字符串，令IsPal(i,j)表示s(i,j)是否为回文字符串（IsPal(i,j)值为1表示是，0表示否），有IsPal(i,j)=IsPal(i+1,j-1)&& s[i]==s[j]；

令minCut(i)表示S(0,i)的最小分隔数，有minCut(i)=min{minCut(j-1)，j<=i且IsPal(j,i)==1}+1，且minCut(0)=0；

其中minCut各项的求解和IsPal各项的求解都使用了动态规划的思想，因为子问题有重叠，所以降低了算法的时间复杂度。

int minCut(string s)
{
vector> isPal(s.length(), vector(s.length(), 0));
vector minCut(s.length(), 0);

for (int i = 0; i < s.length(); i++)
{
for (int j = 0; j < s.length()-i; j++)
{
if (i == 0)
isPal[j][i+j] = 1;
else if (i == 1)
isPal[j][i+j] = (s[j] == s[i+j]);
else
isPal[j][i+j] = (isPal[j+1][i+j-1] && s[j] == s[i+j]);
}
}

minCut[0] = 0;

for (int i = 1; i < s.length(); i++)
{
int minNum = 65536;

for (int j = 0; j < i+1; j++)
{
if (isPal[j][i] == 1)
{
if (j == 0)
{
minNum = -1;
break;
}
else
{
if (minCut[j-1] < minNum)
minNum = minCut[j-1];
}
}
}

minCut[i] = minNum+1;
}

return minCut[s.length()-1];
}