• [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);
        }
    
    }
    View Code

    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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/jdflyfly/p/3812463.html
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