剑指 offer 25. 合并两个排序的链表

剑指 Offer 25. 合并两个排序的链表

输入两个递增排序的链表，合并这两个链表并使新链表中的节点仍然是递增排序的。

示例1：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

限制：

0 <= 链表长度 <= 1000

注意：本题与主站 21 题相同：https://leetcode-cn.com/problems/merge-two-sorted-lists/

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/he-bing-liang-ge-pai-xu-de-lian-biao-lcof 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Solution 1

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode node = head;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                node.next = l1;
                node = node.next;
                l1 = l1.next;
            } else {
                node.next = l2;
                node = node.next;
                l2 = l2.next;
            }
        }
        if (l1 != null)
            node.next = l1;
        else
            node.next = l2;
        return head.next;
    }
}

Solution 2

头指针

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode head = null;
        ListNode node = head;
        if (l1 == null && l2 == null)
            return null;
        if (l1 == null)
            return l2;
        if (l2 == null)
            return l1;
        if (l1.val < l2.val) {
            head = node = l1;
            l1 = l1.next;
        } else {
            head = node = l2;
            l2 = l2.next;
        }
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                node.next = l1;
                node = node.next;
                l1 = l1.next;
            } else {
                node.next = l2;
                node = node.next;
                l2 = l2.next;
            }
        }
        if (l1 != null)
            node.next = l1;
        else
            node.next = l2;
        return head;
    }
}

Solution 3

递归

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null)
            return l2;
        if (l2 == null)
            return l1;
        ListNode node;
        if (l1.val < l2.val) {
            node = l1;
            node.next = mergeTwoLists(l1.next, l2);
        } else {
            node = l2;
            node.next = mergeTwoLists(l1, l2.next);
        }
        return node;
    }
}

Solution 4

cpp 递归

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution
{
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2)
    {
        ListNode *head = NULL;
        if (l1 == NULL)
            return l2;
        if (l2 == NULL)
            return l1;
        if (l1->val <= l2->val)
        {
            head = l1;
            l1->next = mergeTwoLists(l1->next, l2);
        }
        else
        {
            head = l2;
            l2->next = mergeTwoLists(l1, l2->next);
        }
        return head;
    }
};

Solution 5

cpp 循环

class Solution
{
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2)
    {
        ListNode *head = NULL, *tail = NULL, *temp = NULL;
        while (l1 != NULL && l2 != NULL)
        {
            if (l1->val <= l2->val)
            {
                temp = l1;
                l1 = l1->next;
            }
            else
            {
                temp = l2;
                l2 = l2->next;
            }
            if (head == NULL)
            {
                head = tail = temp;
            }
            else
            {
                tail->next = temp;
                tail = temp;
            }
        }
        if (l1 == NULL)
        {
            temp = l2;
        }
        else //if (l2 == NULL)
        {
            temp = l1;
        }
        if (head == NULL)
        {
            head = temp;
        }
        else
        {
            tail->next = temp;
        }
        return head;
    }
};