leetcode第21题——*Merge Two Sorted Lists

题目

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.

思路

把两个有序链表归并成一个有序链表。先用比较笨但很简洁的办法：遍历两个有序链表的节点，比较节点值的大小，将较小值放入目标链表中，遍历完后如果两个链表有剩余节点，由于已经是有序的，插入目标链表的最后那个节点即可。

代码

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
"""
head = ListNode(0)
cur = head

while(l1 != None and l2 != None):
#遍历l1和l2的节点并比较大小，将小的放入目标链表中，cur为游标
if(l1.val < l2.val):
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur.next.next = None
cur = cur.next

#比较完l1和l2后还有剩余节点
if(l1 != None):
cur.next = l1
else:
cur.next = l2

return head.next

Java

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head= new ListNode(0);
ListNode cur = head;
//遍历l1和l2的节点并比较大小，cur代表当前节点
while(l1 != null && l2 != null) {
if(l1.val < l2.val) {
cur.next = l1;
l1 = l1.next;
}
else {
cur.next = l2;
l2 = l2.next;
}
cur.next.next = null;//可减少链表大小，提高效率
cur = cur.next;
}

//比较完l1和l2的节点后还有剩余节点
if(l1 != null) cur.next = l1;
else cur.next = l2;

return head.next;
}
}