题目:原题链接(简单)
标签:链表、双指针
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N ) | O ( 1 ) | 32ms (96.84%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一:
class Solution:
def kthToLast(self, head: ListNode, k: int) -> int:
i1, i2 = head, head
for _ in range(k - 1):
i1 = i1.next
while i1.next:
i1 = i1.next
i2 = i2.next
return i2.val