Leetocde: 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.

方法一DFS: Time Limit Exceeded

void dfs(string s, int depth, int &Min)
	{
		if(s.size() < 1)
		{
			if(Min > depth-1) Min = depth-1;
			return;
		}
		for(int i = s.size()-1; i >= 0; i--)
		{
			if(s[i] != s[0])continue;
			int begin = 0;
			int end = i;
			while(begin < end)
			{
				if(s[begin] == s[end])
				{
					begin++;
					end--;
				}
				else
					break;
			}
			if(begin >= end)
				dfs(s.substr(i+1), depth+1, Min);
		}
	}
	int minCut(string s) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int Min = s.size();
		vector path;
		dfs(s,0,Min);
		return Min;
    }

方法二DP: Accepted

定义函数
D[i,n] = 区间[i,n]之间最小的cut数，n为字符串长度
 a   b   a   b   b   b   a   b   b   a   b   a
                     i                                  n
如果现在求[i,n]之间的最优解？应该是多少？简单看一看，至少有下面一个解
 a   b   a   b   b   b   a   b   b   a   b   a
                    i                  j   j+1      n
此时  D[i,n] = min(D[i, j] + D[j+1,n])  i<=j
D[i] = 区间[i,n]之间最小的cut数，n为字符串长度， 则,
D[i] = min(1+D[j+1] )    i<=j
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]));

int minCut(string s) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int len = s.size();
		int* dp = new int[len+1];
		for(int i=len; i>=0; i--)
			dp[i] = len-i;
		bool** matrix = new bool*[len];
		for(int i=0; i=0; i--)
			for(int j=i; j