leetcode merge list and add two number,ismatch

preview:
‘.‘ Matches any single character.
‘*‘ Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

class Solution {
public:
    bool isMatch(string s, string p) {
        if (p.empty())    return s.empty();

        if (‘*‘ == p[1])
            // x* matches empty string or at least one character: x* -> xx*
            // *s is to ensure s is non-empty
            return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || ‘.‘ == p[0]) && isMatch(s.substr(1), p));
        else
            return !s.empty() && (s[0] == p[0] || ‘.‘ == p[0]) && isMatch(s.substr(1), p.substr(1));
    }
};

class Solution {
public:
    bool isMatch(string s, string p) {
        /**
         * f[i][j]: if s[0..i-1] matches p[0..j-1]
         * if p[j - 1] != ‘*‘
         *      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]
         * if p[j - 1] == ‘*‘, denote p[j - 2] with x
         *      f[i][j] is true iff any of the following is true
         *      1) "x*" repeats 0 time and matches empty: f[i][j - 2]
         *      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]
         * ‘.‘ matches any single character
         */
        int m = s.size(), n = p.size();
        vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false));

        f[0][0] = true;
        for (int i = 1; i <= m; i++)
            f[i][0] = false;
        // p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is ‘*‘ and p[0..j - 3] matches empty
        for (int j = 1; j <= n; j++)
            f[0][j] = j > 1 && ‘*‘ == p[j - 1] && f[0][j - 2];

        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                if (p[j - 1] != ‘*‘)
                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || ‘.‘ == p[j - 1]);
                else
                    // p[0] cannot be ‘*‘ so no need to check "j > 1" here
                    f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || ‘.‘ == p[j - 2]) && f[i - 1][j];

        return f[m][n];
    }
};
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if(l1 == NULL) return l2;
        if(l2 == NULL) return l1;

        if(l1->val < l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwoLists(l2->next, l1);
            return l2;
        }
    }
};
This solution is not a tail-recursive, the stack will overflow while the list is too long :)
http://en.wikipedia.org/wiki/Tail_call

ListNode * addtwonumber(ListNode *l1,ListNode *l2)
{
	ListNode prenode(0),*p = &prenode;
	int extra = 0;
	while (l1 || l2|| extra)
	{
		int sum =(l1?l1->val:0) +(l2?l2->val:0) + extra;
		extra = sum / 10;
		p->next =new ListNode(sum %10);
		l1 = l1 ? l1->next : l1;
		l2 = l2 ? l2->next : l2;
	}
	return prenode.next;
} 

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
{
	ListNode myList(INT_MIN);
	ListNode *p = &myList;
	while (l1 && l2)
	{
		if (l1->val < l2->val)
		{
			p->next=l1;
			l1=l1->next;
		}else
		{
			p->next=l2;
			l2 = l2->next;
		}
		p = p->next;
	}
	p->next = l1 ?l1:l2;
	return p->next;
}