[leetcode] Merge Two Sorted Lists

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.

https://oj.leetcode.com/problems/merge-two-sorted-lists/

 

思路：增加一个dummy head方便处理。然后按照merge的操作来即可。

/**
 * 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) {
        if (l1 == null)
            return l2;
        if (l2 == null)
            return l1;
        ListNode p1 = l1;
        ListNode p2 = l2;

        ListNode head = new ListNode(-1);
        ListNode p = head;

        while (p1 != null && p2 != null) {
            if (p1.val <= p2.val) {
                p.next = p1;
                p1 = p1.next;
            } else {
                p.next = p2;
                p2 = p2.next;
            }
            p = p.next;
        }
        if (p1 != null)
            p.next = p1;
        if (p2 != null)
            p.next = p2;

        return head.next;

    }

    public static void main(String[] args) {
        ListNode l1 = ListUtils.makeList(2, 3, 5, 7);

        ListNode l2 = ListUtils.makeList(1, 4, 5, 9);

        ListNode newHead = new Solution().mergeTwoLists(l1, l2);
        ListUtils.printList(newHead);
    }

}

 

第二遍记录：不小心漏掉了p=p.next

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1==null)
            return l2;
        if(l2==null)
            return l1;
        ListNode dummyHead = new ListNode(-1);
        ListNode p =dummyHead;
        while(l1!=null&&l2!=null){
            if(l1.val<=l2.val){
                p.next =l1;
                l1=l1.next;
            }
            else{
                p.next=l2;
                l2=l2.next;
            }
            p=p.next;
        }
        if(l1!=null)
            p.next=l1;
        if(l2!=null)
            p.next=l2;
        
        return dummyHead.next;
    }
}