Description

  • Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example

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Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

Interface

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/**
 * 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) {
        
    }
};

Solution

合并两个已按小到大排序的链表，最简单的做法就是新建一个链表，然后比较原有的两个链表当前节点的大小，然后选择较小的节点加到新建链表上，接着将被选中的链表的节点指向他的下一个，重复以上，知道某个链表已读到末尾或两个链表都为空，然后将未读完的链表的节点直接加到新建链表上就可以。

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/**
 * 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;
        if (l2 == NULL) return l1;
        ListNode* head = new ListNode(0);
        ListNode* node = head;
        while (l1 != NULL && l2 != NULL) {
            if (l1->val < l2->val) {
                node->next = l1;
                l1 = l1->next;
            } else {
                node->next = l2;
                l2 = l2->next;
            }
            node = node->next;
        }
        node->next = (l1 == NULL ? l2 : l1);
        node = head;
        head = head->next;
        delete node;
        return head;
    }
};

Improvement

还可以通过递归实现，原理与上面相同。

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