Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists.
Example 1:
Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]
Example 2:
Input: l1 = [], l2 = []
Output: []
Example 3:
Input: l1 = [], l2 = [0]
Output: [0]
Constraints:
The number of nodes in both lists is in the range [0, 50].
-100 <= Node.val <= 100
Both l1 and l2 are sorted in non-decreasing order.这道题是经典的考察链表的题目。
l1 和 l2,当 l1 为空或 l2 为空时结束l1 的 val 值更小,则将 l1.next 与排序好的链表头相接,l2 同理O(m+n)O(m+n),mm 为 l1的长度,nn 为 l2 的长度//Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
if l1 == nil {
return l2;
}else if l2 == nil {
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;
}
}执行结果:
leetcode-cn:
执行用时:0 ms, 在所有 Go 提交中击败了100.00%的用户
内存消耗:2.6 MB, 在所有 Go 提交中击败了26.43%的用户
leetcode:
Runtime: 0 ms, faster than 100.00% of Go online submissions for Merge Two Sorted Lists.
Memory Usage: 2.6 MB, less than 51.09% of Go online submissions for Merge Two Sorted Lists.可以看到,该解法执行用时为0ms,非常高效。