[LeetCode] 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.

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题意: 给两个已经按从小到大排好序的整数数组, 要求将这两个数组按从小到大排序成一个新的数组.

思路:

1:若采用给新的数组重新分配内存的方法,这道题思路非常简单,这里不详述;

2:这里作者采用的是不重新分配内存的办法,以原有节点为单元,改变指针的指向,重新排序.

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2)
{
struct ListNode *pL1 = l1;
struct ListNode *pL2 = l2;

struct ListNode *head;  //!< 指向新链表的首节点
struct ListNode *predNode; //!< 指向新链表当前的最后一个节点

if(NULL == pL1 || NULL == pL2)
{
return pL1 == NULL ? pL2:pL1;
}
//!< 判定head节点,并初始化新链表的第一个节点
if(pL1->val  <= pL2->val)
{
head = pL1;
predNode = pL1;
pL1 = pL1->next;
}
else
{
head = pL2;
predNode = pL2;
pL2 = pL2->next;
}

while(1)
{
if( NULL==pL1  || NULL== pL2)
{ //!< 链表生成结束
predNode->next = (pL1==NULL) ? pL2 : pL1;
return head;
}
//!< 判定新的节点
if(pL1->val <= pL2->val)
{
predNode->next = pL1;
predNode = pL1;
pL1 = pL1->next;
}
else
{
predNode->next = pL2;
predNode = pL2;
pL2 = pL2->next;
}
}
//!< 不会执行这一步
return head;
}