*21. Merge Two Sorted Lists[Easy]

21. Merge Two Sorted Lists

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.

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.

Example

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

思路

将两个链表头节点相比较，依次遍历，谁小就尾插进结果节点

题解

   public ListNode mergeTwoLists(ListNode list1, ListNode list2) {

        // 存放结果
        ListNode result = new ListNode(0, new ListNode());
        // 存放结果头节点
        ListNode temp = result;

        // 两个链表只要有一个到头就结束
        while (list1 != null && list2 != null) {
            // 链表1当前节点比链表2大，那就把链表2当前节点续在结果节点后面
            if (list1.val > list2.val) {
                temp.next = new ListNode(list2.val, new ListNode());
                // 遍历下一个节点
                list2 = list2.next;
            } else {
                // 链表1小就把链表1续在结果节点后面
                temp.next = new ListNode(list1.val, new ListNode());
                list1 = list1.next;
            }
            // 结果节点往下移
            temp = temp.next;
        }

        // 最后无非就是把先把链表1遍历完了，或者是先把链表2遍历完了，while就停止=
        // 那这边就看当前哪个链表为null，那么另一个链表就还有值，把其续在结果节点后面
        if (list1 == null)
            temp.next = list2;
        else
            temp.next = list1;

        return result.next;
    }