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 = Noneclass 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;    }}