题21、合并两个有序链表

一、题目¹

二、思路

三、代码

public class T0021 {

    public static void main(String[] args) {

        int[] nums1 = { 1, 3, 5 };
        int[] nums2 = { 2, 4, 6 };

        ListNode l1 = new ListNode(nums1[0]);
        ListNode node = l1;
        for ( int i = 1; i < nums1.length; i++ ){
            node.next = new ListNode(nums1[i]);
            node = node.next;
        }

        ListNode l2 = new ListNode(nums2[0]);
        node = l2;
        for ( int i = 1; i < nums2.length; i++ ){
            node.next = new ListNode(nums2[i]);
            node = node.next;
        }

        ListNode head = mergeTwoLists( l1, l2 );

        while ( head != null ){
            System.out.println( head.val );
            head = head.next;
        }

    }

    public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {

        ListNode head = new ListNode(0);        //头结点
        ListNode node = head;                      //用于迭代的节点
        
        //只要有剩余的节点就继续进行循环
        while ( l1 != null || l2 != null ){
            
            //如果两个节点均不为空，将较小的节点进行保存，较大的不管
            if ( l1 != null && l2 != null ){
                if ( l1.val < l2.val ){
                    node.next = l1;
                    l1 = l1.next;
                }else {
                    node.next = l2;
                    l2 = l2.next;
                }
            //仅l1不为空，将l1保存后退出
            }else if ( l1 != null && l2 == null ){
                node.next = l1;
                break;
            //仅l2不为空，将l2保存后退出 
            }else{
                node.next = l2;
                break;
            }
            
            node = node.next;
        }
        
        //头结点的下一个节点才是有效值
        return head.next;

    }
}
//Definition for singly-linked list.
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) { val = x; }
}

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1. 来源：力扣（LeetCode）
   链接：https://leetcode-cn.com/problems/merge-two-sorted-lists
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