题目地址:leetcode-cn.com/problems/me…
思路:每次取出两个链表中的第一个元素,进行比较。将较小的那个元素插入新链表。一直到某一个链表走完以后,将剩余链表的剩余值插到新链表上。
上代码:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode headNode = new ListNode(0);
ListNode newListNode = headNode;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
newListNode.next = list1;
list1 = list1.next;
} else {
newListNode.next = list2;
list2 = list2.next;
}
newListNode = newListNode.next;
}
newListNode.next = list1 == null?list2:list1;
return headNode.next;
}
}