Problem Description：
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.
分析：设cut[i] = 区间[0，i]之间最小的cut数，n为字符串长度， 则,
cut[i] = min(cut[i]，1+cut[j] ) 0<=j
有个转移函数之后，一个问题出现了，就是如何判断[j,i]是否是回文？每次都从i到j比较一遍？太浪费了，这里也是一个DP问题。
定义函数
flag[i][j] = true if [i,j]为回文
那么
flag[i][j] = str[i] == str[j] && P[i+1][j-1];
class Solution {
public:
int minCut(string s) {
//if(s.size()==0)
// return 0;
int n=s.size();
vector
> flag(n,vector
(n,0)); vector
cut(n+1); for(int i=0;i<=n;i++) cut[i]=i-1; for(int i=0;i