LeetCode 019. Remove Nth Node From End of List (Easy)

给出一个链表,移除其中倒数第N个元素。

原始问题

https://leetcode.com/problems/remove-duplicates-from-sorted-list/

Given a linked list, remove the nth node from the end of list and return its head.

For example,
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

解题思路

双指针,后一个指针比前一个指针先走N步,后一个指针移动到链表末尾时前一个指针指向的就是要删除的元素。

AC代码

C语言

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struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
struct ListNode * p1, * p2, * tmp;
struct ListNode dummy;

dummy.next = head;
p1 = &dummy;
p2 = &dummy;

for (int i = 0; i < n; i++) {
p2 = p2->next;
}

while (p2->next != NULL) {
p1 = p1->next;
p2 = p2->next;
}

tmp = p1->next;
p1->next = p1->next->next;
free(tmp);

return dummy.next;
}

这里使用了正确的删除节点的方法,使用free()释放了内存空间。

文章目录
  1. 1. 原始问题
  2. 2. 解题思路
  3. 3. AC代码
    1. 3.1. C语言
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