leetcode-21-链表操作

/**
<p>将两个升序链表合并为一个新的 <strong>升序</strong> 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。 </p>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/merge_ex1.jpg" style="width: 662px; height: 302px;" />
<pre>
<strong>输入：</strong>l1 = [1,2,4], l2 = [1,3,4]
<strong>输出：</strong>[1,1,2,3,4,4]
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>l1 = [], l2 = []
<strong>输出：</strong>[]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>l1 = [], l2 = [0]
<strong>输出：</strong>[0]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>两个链表的节点数目范围是 <code>[0, 50]</code></li>
	<li><code>-100 <= Node.val <= 100</code></li>
	<li><code>l1</code> 和 <code>l2</code> 均按 <strong>非递减顺序</strong> 排列</li>
</ul>
<div><div>Related Topics</div><div><li>递归</li><li>链表</li></div></div><br><div><li>👍 2323</li><li>👎 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * 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 p = dummy;
		while(list1!=null && list2!=null){
			if(list1.val<list2.val){
				p.next=list1;
				list1=list1.next;
			}else{
				p.next=list2;
				list2=list2.next;
			}
			p=p.next;
		}
		if(list1!=null){
			p.next=list1;
		}
		if(list2!=null){
			p.next=list2;
		}

		return dummy.next;
    }
}
//leetcode submit region end(Prohibit modification and deletion)