Leetcode75 Day3

第一题

暴力解法：

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

            // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
            prev.next = l1 == null ? l2 : l1;

            return prehead.next;

        }
    }

递归

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;
        }
    }

}

第二题：

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}