链表问题_合并两个排序链表

题目：输入两个递增排序的链表，合并这两个排序链表，使得合并之后的链表仍然使递增排序的。
分析：即是归并排序中的归并操作，有两种具体的实现：1、使用递归的方法，2、迭代版本，逐个合并链表。实现如下：

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/

/*递归版本*/
  ListNode*merge(ListNode*phead1, ListNode*phead2)
  {
      if (phead1 == NULL)
          return phead2;
      else if (phead2 == NULL)
          return phead1;

      ListNode*pmergehead = NULL;
      if (phead1->val < phead2->val)
      {
          pmergehead = phead1;
          pmergehead->next=merge(phead1->next, phead2);
      }
      else
      {
          pmergehead = phead2;
          pmergehead->next=merge(phead1, phead2->next);
      }
     return pmergehead ;
  }

/*迭代版本*/
ListNode*Merge(ListNode*phead1, ListNode*phead2)
  {
      if (phead1 == NULL)
          return phead2;
      else if (phead2 == NULL)
          return phead1;

      ListNode* pnode1 = phead1;
      ListNode* pnode2 = phead2;
      ListNode* pmergehead = NULL;

      if (phead1->val < phead2->val)
      {
          pnode1 = pnode1->next;
          pmergehead = phead1;
      }
      else
      {
          pnode1 = pnode1->next;
          pmergehead = phead1;
      }
      ListNode*pnode3 = pmergehead;
      while (pnode1&&pnode2)
      {
          if (pnode1->val < pnode2->val)
          {
              pnode3->next = pnode1;
              pnode3 = pnode3->next;
              pnode1 = pnode1->next;
          }
          else
          {
              pnode3->next = pnode2;
              pnode3 = pnode3->next;
              pnode2 = pnode2->next;
          }
      }

      while (pnode1)
      {
          pnode3->next = pnode1;
          pnode3 = pnode3->next;
          pnode1 = pnode1->next;
      }

      while (pnode2)
      {
          pnode3->next = pnode2;
          pnode3 = pnode3->next;
          pnode2 = pnode2->next;
      }
      return pmergehead;
  }