合并两个有序链表

原题链接

递归（要开栈，占空间）

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

迭代

设置一个dummy做头，接下来操作链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
 class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode curNode = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                curNode.next = l1;
                l1 = l1.next;
            } else {
                curNode.next = l2;
                l2 = l2.next;
            }
            curNode = curNode.next;
        }
        curNode.next = l1 == null ? l2 : l1;
        return dummy.next;
    }
}