力扣题目学习笔记(OC + Swift)21. 合并两个有序链表

21. 合并两个有序链表

将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

链表解题经典三把斧：

• 哑巴节点
• 栈
• 快慢指针

此题比较容易想到的解法是迭代法，生成哑巴节点，然后迭代生成后续节点。

方法一、迭代法

Swift

func mergeTwoLists(_ list1: ListNode?, _ list2: ListNode?) -> ListNode? {guard list1 != nil else {return list2}guard list2 != nil else {return list1}var list1 = list1var list2 = list2let dummyNode = ListNode(-1);var prev:ListNode? = dummyNodewhile list1 != nil && list2 != nil {if list1!.val < list2!.val {prev?.next = list1list1 = list1!.next}else {prev?.next = list2list2 = list2!.next}prev = prev?.next}prev?.next = (list1 != nil) ? list1 : list2return dummyNode.next}

OC

//回溯法
- (ListNodeOC *_Nullable)mergeTwoLists:(ListNodeOC * _Nullable)list1list2:(ListNodeOC * _Nullable)list2 {if (!list1) {return list2;}if (!list2) {return list1;}ListNodeOC *dummyNode = [[ListNodeOC alloc] initWithVal:-1];ListNodeOC *pre = dummyNode;while (list1 && list2) {if (list1.val < list2.val) {pre.next = list1;list1 = list1.next;}else {pre.next = list2;list2 = list2.next;}pre = pre.next;}pre.next = list1 ? list1 : list2;return dummyNode.next;
}

方法二、递归法

代码简洁、思路清晰、稍占内存的解法。

Swift

func mergeTwoLists(_ list1: ListNode?, _ list2: ListNode?) -> ListNode? {guard let list1 = list1 else { return list2 }guard let list2 = list2 else { return list1 }if list1.val < list2.val {list1.next = mergeTwoLists(list1.next, list2)return list1}else {list2.next = mergeTwoLists(list1, list2.next)return list2}}

OC

//递归法
- (ListNodeOC * _Nullable)mergeTwoLists:(ListNodeOC * _Nullable)list1list2:(ListNodeOC * _Nullable)list2 {//递归终止条件if (!list1) {return list2;}if (!list2) {return list1;}if (list1.val < list2.val) {list1.next = [self mergeTwoLists:list1.next list2:list2];return list1;}else {list2.next = [self mergeTwoLists:list1 list2:list2.next];return list2;}
}