题目描述
🔥 21. 合并两个有序链表
思路分析
这道题目是链表的基础题目，我们可以使用递归或者迭代的方式来实现。
参考代码
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| func mergeTwoLists(list1 *ListNode, list2 *ListNode) *ListNode {
if list1 == nil {
return list2
} else if list2 == nil {
return list1
}
if list1.Val < list2.Val {
list1.Next = mergeTwoLists(list1.Next, list2)
return list1
} else {
list2.Next = mergeTwoLists(list1, list2.Next)
return list2
}
}
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| func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
if l1 == nil {
return l2
} else if l2 == nil {
return l1
}
dummy := &ListNode{}
cur := dummy
for l1 != nil && l2 != nil {
if l1.Val < l2.Val {
cur.Next = l1
l1 = l1.Next
} else {
cur.Next = l2
l2 = l2.Next
}
cur = cur.Next
}
if l1 != nil {
cur.Next = l1
}
if l2 != nil {
cur.Next = l2
}
return dummy.Next
}
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🍏 点击查看 Java 题解
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| class ListNode {
int val;
ListNode next;
ListNode() {
}
ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode();
ListNode cur = dummy;
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;
}
if (l1 != null) {
cur.next = l1;
}
if (l2 != null) {
cur.next = l2;
}
return dummy.next;
}
}
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| class ListNode {
int val;
ListNode next;
ListNode() {
}
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;
}
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}
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