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 ofs.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

DP.

res[i] 区间[i,n]之间最小的cut数，n为字符串长度， 则,

res[i] = min(1+res[j+1], res[i] )    i<=j <n.

有个转移函数之后，一个问题出现了，就是如何判断[i,j]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
P[i][j] = true if [i,j]为回文

那么
P[i][j] = (str[i] == str[j] && P[i+1][j-1]);

class Solution {public:     int minCut(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int n = s.size();        vector<int > res(n+1);        vector<vector<bool> > p(n, vector<bool>(n, false));        for(int i = 0 ;i <= n; ++i){            res[i] = n - i;        }        for(int i = n - 1; i >=0; --i){            for(int j = i; j < n; ++j){                if(s[i] == s[j] && (j - i < 2 || p[i + 1][j - 1])){                    p[i][j] = true;                    res[i] = min(res[i], res[j + 1] + 1);                }            }        }        return res[0] - 1;    }};

也可以从前往后数。

class Solution {public:    int minCut(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int n = s.size();        vector<int> res(n+1);        for(int i = 0 ; i< n + 1 ; ++i){            res[i] = i - 1;        }        vector<vector<bool> > p(n, vector<bool>(n, false));        for(int i = 0; i< n; ++i){            for(int j = 0; j <= i ; ++j){                if(s[i] == s[j] && (i - j < 2 || p[j + 1][i - 1])){                    p[j][i] = true;                    res[i + 1] = min(res[i + 1], res[j] + 1);                }            }        }        return res[n];    }};