Merge Two Sorted Listsf

Description

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Example 1:

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = []
Output: []

Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

Constraints:

• The number of nodes in both lists is in the range [0, 50].

• -100 <= Node.val <= 100

• Both list1 and list2 are sorted in non-decreasing order.

Solution

Obviously, this is problem regarded as the combine of many smaller problems, solved using a typical recursive algorithm.

1) Finding the basic cases

• Given whether both linked lists are empty

• Given whether any linked list is empty

2) Finding the recursive relationship

• After compare the first element, according to the compare result, to decide next round compare item. E.g. The first element of list1 is smaller than the first element of list2, the next round will compare the second element of list1 and first element of list2. Otherwise, will compare the first element of list1 and second element of list2.

• The result of each round comparion will combine a linked list.

/* Definition for singly-linked list.
 * public class ListNode {
 *      int val;
 *      ListNode next;
 *      ListNode() {}
 *      ListNode(int val) { this.val = val; }
 *      ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 *  }
*/

class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if (list1 != null && list2 != null) {
            if (list1.val <= list2.val) {
                list1.next = mergeTwoLists(list1.next, list2);
                return list1;
            } else {
                list2.next = mergeTwoLists(list1, list2.next);
                return list2;
            }
        }
        if (list1 == null)
            return list2;
        return list1;
    }
}

Time Complexity

O(n)