LeetCode

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.

Difficulty

Solution

Idea

由于两个列表已排好序，因此可以设置两个下标i，j，分别指向l1和l2的首节点，再从头到尾分别比较两个列表当前下标对应的元素的大小关系，接着存到新列表（注意，这里题干要求“return it as a new list”）。需要留意两个列表的长度不一致的情况。

python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if l1 == None and l2 == None:
            return None

        result = ListNode(-1)
        result_cur = result

        ptr1 = l1
        ptr2 = l2

        while ptr1 and ptr2:
            if ptr1.val <= ptr2.val:
                new_node = ListNode(ptr1.val)
                result_cur.next = new_node
                ptr1 = ptr1.next
            else:
                new_node = ListNode(ptr2.val)
                result_cur.next = new_node
                ptr2 = ptr2.next

            result_cur = result_cur.next

        while ptr1:
            new_node = ListNode(ptr1.val)
            result_cur.next = new_node
            result_cur = result_cur.next
            ptr1 = ptr1.next

        while ptr2:
            new_node = ListNode(ptr2.val)
            result_cur.next = new_node
            result_cur = result_cur.next
            ptr2 = ptr2.next

        return result.next

cpp

/**
 * 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 && l2==NULL)
        {
            return NULL;
        }
        ListNode* ptr1 = l1;
        ListNode* ptr2 = l2;
        ListNode* result = new ListNode(-1);
        ListNode* result_cur = result;

        while (ptr1 && ptr2)
        {
            if (ptr1->val <= ptr2->val)
            {
                result_cur->next = new ListNode(ptr1->val);;
                ptr1 = ptr1->next;
            }else
            {
                result_cur->next = new ListNode(ptr2->val);
                ptr2 = ptr2->next;
            }
            result_cur = result_cur->next;
        }
        while (ptr1)
        {
            result_cur->next = new ListNode(ptr1->val);
            result_cur = result_cur->next;
            ptr1 = ptr1->next;
        }
        while (ptr2)
        {
            result_cur->next = new ListNode(ptr2->val);;
            result_cur = result_cur->next;
            ptr2 = ptr2->next;
        }

        return result->next;
    }
};