LeetCode 19. 删除链表的倒数第 N 个结点
题目描述
思路分析
快慢指针问题
参考代码
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func removeNthFromEnd(head *ListNode, n int) *ListNode {
dummy := &ListNode{Next: head}
slow, fast := dummy, dummy
for i := 0; i < n; i++ {
fast = fast.Next
}
for fast != nil && fast.Next != nil {
fast = fast.Next
slow = slow.Next
}
slow.Next = slow.Next.Next
return dummy.Next
}
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func removeNthFromEnd(head *ListNode, n int) *ListNode {
dummy := &ListNode{Next: head}
slow, fast := dummy, dummy
// 让 fast 指针先走 n 步
for i := 0; i <= n; i++ {
fast = fast.Next
}
// fast 和 slow 指针一起走，直到 fast 到达链表末尾
for fast != nil {
slow = slow.Next
fast = fast.Next
}
// 删除倒数第 N 个节点
slow.Next = slow.Next.Next
return dummy.Next
}
- 时间复杂度：O(n)，其中
n是链表的节点数。- 空间复杂度：O(1)，只使用了常数级别的空间。
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class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode();
dummy.next = head;
ListNode slow = dummy, fast = dummy;
for (int i = 0; i < n; i++) {
fast = fast.next;
}
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next;
return dummy.next;
}
}
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