[leetCode]剑指 Offer 25. 合并两个排序的链表

递归

可以逐一比较两个链表的头节点，将交小的节点取出放到链表尾部。当链表1为空时返回链表2，当链表2为空时，返回链表1。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null) return l2;
        else if(l2 == null) return l1;

        ListNode pMergeHead = null;

        if(l1.val >= l2.val){
            pMergeHead = l2;
            l2.next = mergeTwoLists(l1,l2.next);
        }else{
            pMergeHead = l1;
            l1.next = mergeTwoLists(l1.next,l2);
        }
        return pMergeHead;
    }
}

非递归写法，使用哑节点

两种写法思路一样，第二种简洁

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummyNode = new ListNode();
        ListNode p1 = l1;
        ListNode p2 = l2;
        ListNode p = dummyNode;
        while(p1 != null || p2 != null){
            if(p1 == null){
                p.next = p2;
                break;
            }else if(p2 == null){
                p.next = p1;
                break;
            }
            if(p1.val <= p2.val ){
                p.next = p1;
                p1 = p1.next;
                p = p.next;
            }else {
                p.next = p2;
                p2 = p2.next;
                p =p.next;
            }
        }
        return dummyNode.next;
    }
}

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummyNode = new ListNode();
        ListNode cur = dummyNode;
        while(l1 != null && l2 != null){
            if(l1.val <= l2.val ){
                cur.next = l1;
                l1 = l1.next;
            }else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1 == null ? l2 : l1;
        return dummyNode.next;
    }
}