Merge Two Sorted Lists

Merge Two Sorted Lists 

问题：

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.

思路：

　　归并排序的后续处理方式

我的代码：

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) {

 *         val = x;

 *         next = null;

 *     }

 * }

 */

public class Solution {

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {

        ListNode dummy = new ListNode(-1);

        ListNode rst = dummy;

        while(l1 != null && l2 != null)

        {

            int val1 = l1.val;

            int val2 = l2.val;

            if(val1 < val2)

            {

                dummy.next = l1;

                l1 = l1.next;

            }

            else

            {

                dummy.next = l2;

                l2 = l2.next;

            }

            dummy = dummy.next;

        }

        while(l1 != null)

        {

            dummy.next = l1;

            l1 = l1.next;

            dummy = dummy.next;

        }

        while(l2 != null)

        {

            dummy.next = l2;

            l2 = l2.next;

            dummy = dummy.next;

        }

        return rst.next;

    }

}

View Code

他人代码：

public class Solution {

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {

        ListNode dummy = new ListNode(0);

        ListNode lastNode = dummy;

        

        while (l1 != null && l2 != null) {

            if (l1.val < l2.val) {

                lastNode.next = l1;

                l1 = l1.next;

            } else {

                lastNode.next = l2;

                l2 = l2.next;

            }

            lastNode = lastNode.next;

        }

        

        if (l1 != null) {

            lastNode.next = l1;

        } else {

            lastNode.next = l2;

        }

        

        return dummy.next;

    }

}

View Code

学习之处：

• 链表不同之处在于最后做收尾工作的时候，不用再继续循环了，直接if else就OK了，数组不同于链表啊，我的代码中在这一部分浪费了很多时间