【CT】LeetCode手撕—21. 合并两个有序链表

题目

• 原题连接：21. 合并两个有序链表

1-思路

双指针：题目提供的 list1 和 list2 就是两个双指针

• 通过每次移动 list1 和 list2 并判断二者的值，判断完成后将其 插入到新的当前结点 cur.next 后即可完成链表按升序排列。

2- 实现

⭐21. 合并两个有序链表——题解思路

class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        // 定义一个 头结点记录位置
        ListNode dummyHead = new ListNode(0);
        ListNode cur = dummyHead;
        while(list1!=null && list2!=null){
            if(list1.val<list2.val){
                cur.next = list1;
                list1 = list1.next;
            }else{
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }

        if(list1!=null){
            cur.next = list1;
        }
        if(list2!=null){
            cur.next = list2;
        }
        return dummyHead.next;
    }
}

3- ACM实现

public class twoLinkASC {

    static class ListNode{
        int val;
        ListNode next;
        ListNode(){}
        ListNode(int x){
            val = x;
        }
    }

    public static ListNode mergeTwoLink(ListNode list1,ListNode list2){
        // 定义虚拟头
        ListNode dummyHead = new ListNode(-1);
        ListNode cur = dummyHead;

        while(list1!=null && list2!=null){
            if(list1.val<list2.val){
                cur.next = list1;
                list1 = list1.next;
            }else{
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }
        if(list1!=null){
            cur.next = list1;
        }
        if(list2!=null){
            cur.next = list2;
        }
        return dummyHead.next;
    }


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("输入链表1的长度");
        ListNode head1=null,tail1=null;
        int n1 = sc.nextInt();
        System.out.println("输入链表元素");
        for(int i = 0 ; i < n1;i++){
            ListNode nowNode = new ListNode(sc.nextInt());
            if(head1==null){
                head1 = nowNode;
                tail1 = nowNode;
            }else{
                tail1.next = nowNode;
                tail1 = nowNode;
            }
        }

        System.out.println("输入链表2的长度");
        ListNode head2=null,tail2=null;
        int n2 = sc.nextInt();
        System.out.println("输入链表元素");
        for(int i = 0 ; i < n2;i++){
            ListNode nowNode = new ListNode(sc.nextInt());
            if(head2==null){
                head2 = nowNode;
                tail2 = nowNode;
            }else{
                tail2.next = nowNode;
                tail2 = nowNode;
            }
        }

        ListNode forRes = mergeTwoLink(head1,head2);
        while (forRes!=null){
            System.out.print(forRes.val+" ");
            forRes = forRes.next;
        }
    }
}