Palindrome Partitioning II

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

解法：还比较典型的动态规划，需要注意的是判断子串对偶也用的是动态规划。

 定义子问题：dp[i]表示从i到n的子串划分成的回文串的个数，减1即划分的刀数，palin[i][j]表示子串i...j是否为回文串

递归关系：dp[i] = min(dp[i],dp[j+1]+1); i <=j < n

                    palin[i][j] = pos[i+1][j-1]&&(s[i]==s[j])  || (i == j)

初始条件：dp[i]=n-i;pos[i][j] = false;

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

124 milli secs.