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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → trueclass 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
     
      
       > f(m + 1, vector
       
        (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_callListNode * 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;}