21. Merge Two Sorted Lists

   Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

// recursive solution, time O(m+n), much more space complexity

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */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(l1, l2->next);            return l2;         }                 }};

  my 非递归法

class Solution {public:    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {        if(l1 == NULL)        return l2;        if(l2 == NULL)        return l1;        ListNode* p = l1;        ListNode* q = l2;        ListNode* x = (l1->val <= l2->val) ? l1 :l2;        ListNode* c;                        while(p&&q)        {            if(p->val <=q->val)            {                c->next = p;                c = c->next;                p = p->next;            }            else            {                c->next = q;                c = c->next;                q = q->next;            }                    }        if(!p)        {            c->next = q;                    }        if(!q)        {            c->next = p;                    }        return x;                         }};

// merge sort, adding while comparing, time O(m+n)class Solution_MergeSort {public:    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {        ListNode dummy(0);                                     //头！        ListNode *curr = &dummy;        while (l1 && l2)        {            if (l1->val < l2->val)            {                curr->next = l1;                curr = l1;                l1 = l1->next;            }            else            {                curr->next = l2;                curr = l2;                l2 = l2->next;            }        }        if (!l1)        {            curr->next = l2;        }        if (!l2)        {            curr->next = l1;        }        return dummy.next;    }};