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2014
11-18

LeetCode-Remove Nth Node From End of List[链表]

Remove Nth Node From End of 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.

标签: Linked List Two Pointers
分析

设两个指针$p,q$,让$q$先走$n$步,然后$p$和$q$一起走,直到$q$走到尾节点,删除{p->next}即可。

代码1

// LeetCode, Remove Nth Node From End of List
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode dummy{-1, head};
        ListNode *p = &dummy, *q = &dummy;

        for (int i = 0; i < n; i++)  // q先走n步
            q = q->next;

        while(q->next) { // 一起走
            p = p->next;
            q = q->next;
        }
        ListNode *tmp = p->next;
        p->next = p->next->next;
        delete tmp;
        return dummy.next;
    }
};