Remove Nth Node From End of List

最新推荐文章于 2025-09-09 19:16:33 发布

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最新推荐文章于 2025-09-09 19:16:33 发布

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文章标签： c/c++

原文链接：http://www.cnblogs.com/kwill/p/3189736.html

本文提供了一种高效的方法来移除链表中倒数第N个节点。通过一次遍历实现，避免了多次遍历所带来的额外开销。文章包含完整的C++代码实现。

Given a linked list, remove the n^th 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.

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!head || n <= 0) return head;
        ListNode dummynode(0), *h = &dummynode,*p = head,*e = p;
        h->next = head;
        int k = 0;
        for(;k < n && e; k++){
            e = e->next;
        }
        //n正好为总长度
        if (!e){
            return head->next;
        }
        while(p && e && e->next){
            p = p->next;
            e = e->next;
        }
        if (p) p->next = p->next->next;
        return h->next; 
    }
};