剑指Offer_25_合并两个排序的链表

❗️LeetCode_21_合并两个有序链表

题目描述：

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

限制：0 <= 链表长度 <= 1000

解法1：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) {

        if (l1 == NULL) {
            return l2;
        } else if (l2 == NULL) {
            return l1;
        }

        ListNode *pMergedHead = NULL;

        //递归
        if (l1->val < l2->val) {
            pMergedHead = l1;
            pMergedHead->next = mergeTwoLists(l1->next,l2);
        } else {
            pMergedHead = l2;
            pMergedHead->next = mergeTwoLists(l1,l2->next);
        }

        return pMergedHead;
    }
};

解法2：迭代

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {

        ListNode* preHead = new ListNode(-1);

        ListNode* prev = preHead;

        while (l1 != nullptr && l2 != nullptr) {

            if (l1->val <= l2->val) {
                prev->next = l1;
                l1 = l1->next;
            } else {
                prev->next = l2;
                l2 = l2->next;
            }
            prev = prev->next;
        }

        //合并后l1和l2最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
        //l1现在如果指向空，prev->next指向l2，否则指向l1
        prev->next = l1 == nullptr ? l2 : l1;

        return preHead->next;
    }
};