21. Merge Two Sorted Lists(Linked List-Easy)

转载请注明作者和出处：http://blog.csdn.net/c406495762

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.

将两个排序的链表合并，返回一个新链表，返回的新链表也是排好序的。

解题思路：

创建两个链表，一个负责保存头节点，一个负责记录比较后的结果。

Language : c

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
struct ListNode* newlist = (struct ListNode *)malloc(sizeof(struct ListNode));
struct ListNode* temp = (struct ListNode *)malloc(sizeof(struct ListNode));
newlist = temp;
while(l1 && l2){
if(l1->val < l2->val){
temp->next = l1;
l1 = l1->next;
}
else{
temp->next = l2;
l2 = l2->next;
}
temp = temp->next;
}
temp->next = l1 ? l1 : l2;
return newlist->next;
}

Language : cpp

/**
* 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) {
ListNode newlist(INT_MIN);
ListNode *temp = &newlist;
if(l1 == NULL && l2 == NULL){
return NULL;
}
if(l1 != NULL && l2 == NULL){
return l1;
}
if(l1 == NULL && l2 != NULL){
return l2;
}
while(l1 && l2){
if(l1->val < l2->val){
temp->next = l1;
l1 = l1->next;
}
else{
temp->next = l2;
l2 = l2->next;
}
temp = temp->next;
}
temp->next = l1 ? l1 : l2;
return newlist.next;
}
};

Language：python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
result = cur = ListNode(0)
while l1 and l2:
if l1.val < l2.val:
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = l1 or l2
return result.next

LeetCode题目汇总： https://github.com/Jack-Cherish/LeetCode