package pers.vic.basics.leetcode;

/**
 * @author Vic.xu
 * @description: 21. 合并两个有序链表
 * @date: 2020/11/11 0011 11:36
 */
public class J0021_E_MergeTwoLists {
    /*
    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。
     */

    /**
     * 递归遍历
     */
    public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2==null){
            return l1;
        }

        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        }else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }

    //暴力遍历 反正是有序的
    public static ListNode mergeTwoLists2(ListNode l1, ListNode l2) {
        //暴力遍历：有序的
        ListNode dummy = new ListNode();
        ListNode pref = dummy;
        while (l1 != null && l2 != null) {
                if (l1.val < l2.val) {
                    pref.next = l1;
                    l1 = l1.next;
                } else {
                    pref.next = l2;
                    l2 = l2.next;
                }
            pref = pref.next;
        }
        if (l1 != null) {
            pref.next = l1;
        } else if (l2 != null) {
            pref.next  = l2;
        }
        return dummy.next;
    }


    /*
    输入：1->2->4,    1->3->4
    输出：1->1->2->3->4->4
     */
    public static void main(String[] args) {
        //1->2->4
        ListNode n1 = new ListNode(4);
        ListNode n2 = new ListNode(2, n1);
        ListNode l1 = new ListNode(1, n2);

//        1->3->4
        ListNode n3 = new ListNode(4);
        ListNode n4 = new ListNode(3, n3);
        ListNode l2 = new ListNode(1, n4);
        ListNode head = mergeTwoLists(l1, l2);
        ListNode node = head;
        while (node != null) {
            System.out.println(node.val + "→");
            node = node.next;

        }

    }
}
//  Definition for singly-linked list.