LeetCode 21: 合并两个有序链表

[guodongxiaren]

https://leetcode-cn.com/problems/merge-two-sorted-lists

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。 

示例：

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

[guodongxiaren]

一个叹为观止的解法，利用递归，倒着合并。

递归解法

/**
 * 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) {
            return l2;
        } else if (!l2) {
            return l1;
        } else if (l1->val < l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};

[guodongxiaren]

迭代解法

/**
 * 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 dummyNode;
        ListNode *pNode = &dummyNode;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                pNode->next = l1;
                l1 = l1->next;
            } else {
                pNode->next = l2;
                l2 = l2->next;
            }
            pNode = pNode->next;
        }
        if (l1) {
            pNode->next = l1;
        }
        if (l2) {
            pNode->next = l2;
        }
        return dummyNode.next;
    }
};