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#yyds干货盘点# LeetCode 热题 HOT 100：合并两个有序链表
source link: https://blog.51cto.com/u_13321676/5689371
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#yyds干货盘点# LeetCode 热题 HOT 100：合并两个有序链表
精选 原创将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]
输入：l1 = [], l2 = []
输出：[]
输入：l1 = [], l2 = [0]
输出：[0]
代码实现：
/**
* 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 dummy = new ListNode(-1);
ListNode cur = dummy;
while(list1 != null && list2 != null){
if(list1.val < list2.val){
cur.next = new ListNode(list1.val);
list1 = list1.next;
} else {
cur.next = new ListNode(list2.val);
list2 = list2.next;
}
cur = cur.next;
}
cur.next = list1 == null ? list2 : list1;
return dummy.next;
}
}
* 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 dummy = new ListNode(-1);
ListNode cur = dummy;
while(list1 != null && list2 != null){
if(list1.val < list2.val){
cur.next = new ListNode(list1.val);
list1 = list1.next;
} else {
cur.next = new ListNode(list2.val);
list2 = list2.next;
}
cur = cur.next;
}
cur.next = list1 == null ? list2 : list1;
return dummy.next;
}
}
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