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.
无耻一点的话，可以再问题一的基础上，找结果集中元素最少的，当然就是分割最少的，厚道一些就dp了。
开始觉得很简答，利用问题一中的 判断子串是否是回文串的函数，进行dp，但是果断超时了，看来是否为回文的信息还要记录，就添加个数组，来记录从i 到 j 的字串是否为回文。
class Solution {
public:
/*
dp[k] = min{ dp[k], dp[j - 1] + 1} if s[j...k] is palindrom, 0 <=j <= k - 1;
dp[k] = dp[k - 1] + 1，otherwise.
*/
int minCut(string s){
vector<int> dp(s.size() + 1, 0x7f7f7f7f);
/*
bplin[i][i] = true, but it‘s useless for this problem.
*/
vector<vector<bool> > bpalin(s.size(), vector<bool>(s.size(), false));
dp[0] = -1;
for(int i = 0; i < s.size(); ++i)
{
//we assume s[i] can‘t make palindrome with items before it.
dp[i + 1] = dp[i] + 1;
//check if s[i] can make palindrome with items before it.
//if so, we change dp[i + 1].
for(int cur = i - 1; cur >= 0; --cur)
if(s[i] == s[cur] && (i - cur <= 2 || bpalin[cur + 1][i - 1])){
dp[i + 1] = min(dp[i + 1], dp[cur] + 1);
bpalin[cur][i] = true;
}
}
return dp[s.size()];
}
};