leetcode21- Merge Two Sorted Lists- easy

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.

问清楚需不需要创造新节点。算法比较简单，比较后接上即可，就是小心心里跑一下corner case，看[][], [][1], [1][] 这几个会不会有空指针问题。

1. 创造新节点

/**
 * 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) {
        
        ListNode dummy = new ListNode(-1);
        ListNode crt = dummy;
        while (l1 != null || l2 != null) {
            // 一定要写全条件了，这样写对l1为null但l2不为null的case走进去判断会报空指针！！
            // if (l2 == null || l1.val <= l2.val) {
            if (l1 == null) {
                crt.next = new ListNode(l2.val);
                crt = crt.next;
                l2 = l2.next;
            } else if (l2 == null) {
                crt.next = new ListNode(l1.val);
                crt = crt.next;
                l1 = l1.next;
            } else if (l1.val <= l2.val) {
                crt.next = new ListNode(l1.val);
                crt = crt.next;
                l1 = l1.next;
            } else {
                crt.next = new ListNode(l2.val);
                crt = crt.next;
                l2 = l2.next;
            }
        }
        return dummy.next;
    }
}

2.不创造新节点

/**
 * 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) {
        
        ListNode dummy = new ListNode(-1);
        ListNode crt = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                crt.next = l1;
                l1 = l1.next;
            } else {
                crt.next = l2;
                l2 = l2.next;
            }
            crt = crt.next;
        }
        
        if (l1 == null && l2 != null) {
            crt.next = l2;
        } else if (l1 != null && l2 == null) {
            crt.next = l1;
        }
        return dummy.next;
    }
}