LeetCode 21 — Merge Two Sorted Lists（C++ Java Python）

题目：http://oj.leetcode.com/problems/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.

题目翻译：

合并两个有序链表，并返回该新链表。新链表应该由前两个链表中的节点拼接生成。
分析：

        使用归并的思想，每次选择值较小的节点接到新链表的后面。

C++实现：

/**
* 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;
}

ListNode *node = NULL;

if(l1->val <= l2->val)
{
node = l1;
l1 = l1->next;
}
else
{
node = l2;
l2 = l2->next;
}

ListNode *head = node;

while(l1 != NULL && l2 != NULL)
{
if(l1->val <= l2->val)
{
node->next = l1;
node = node->next;

l1 = l1->next;
}
else
{
node->next = l2;
node = node->next;

l2 = l2->next;
}
}

if(l1 != NULL)
{
node->next = l1;
}
else if(l2 != NULL)
{
node->next = l2;
}

return head;
}
};

Java实现：

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}

ListNode node = null;
ListNode head = null;

while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
if (node == null) {
node = l1;
head = node;
} else {
node.next = l1;
node = node.next;
}

l1 = l1.next;
} else {
if (node == null) {
node = l2;
head = node;
} else {
node.next = l2;
node = node.next;
}

l2 = l2.next;
}
}

if (l1 != null) {
node.next = l1;
} else if (l2 != null) {
node.next = l2;
}

return head;
}
}

Python实现：

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

class Solution:
# @param two ListNodes
# @return a ListNode
def mergeTwoLists(self, l1, l2):
if l1 == None:
return l2
if l2 == None:
return l1

node = None

while l1 != None and l2 != None:
if l1.val <= l2.val:
if node == None:
node = l1
head = node
else:
node.next = l1
node = node.next

l1 = l1.next
else:
if node == None:
node = l2
head = l2
else:
node.next = l2
node = node.next

l2 = l2.next

if l1 != None:
node.next = l1
elif l2 != None:
node.next = l2

return head

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