15 - Remove Nth Node From End of List-优快云博客

本文链接：https://blog.youkuaiyun.com/superrobin/article/details/11597709

本文介绍了一种优化的算法来删除单链表中倒数第N个节点，确保操作在一次遍历内完成，并讨论了算法实现时需要考虑的特殊情况。

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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.

solution: 需考虑如下情况

1： head为空或n<= 0;

2：当n为1时，即删除尾结点（同时考虑如果只存在一个结点）;

3：若n大于整个链表长度，在循环得到距离head为n的结点时需要注意；

4：遍历到倒数第n个结点了，若此结点为头结点，那么需返回head->next；

5：这个时候就可以处理最普遍的情况了，把后一个结点的值覆盖此结点，再删除后一个结点，即可完成删除倒数nth结点操作。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(head == NULL || n <= 0)
            return NULL;
        else if(n == 1)
        {
            if(head -> next == NULL)
                return NULL;
            ListNode* temp = head;
            while(temp->next->next != NULL)
                temp = temp -> next;
            temp->next = NULL;
            return head;
        }
        
        ListNode* nthNode = head;
        int i = 1;
        while(i < n)
        {   
            if(nthNode->next != NULL)
                nthNode = nthNode -> next;
            else
                return NULL;
            i++;
        }
        
        
        ListNode* node = head;
        
        while(nthNode->next != NULL)
        {
            node = node -> next;
            nthNode = nthNode -> next;
        }
        
        if(node == head)
            return head->next;
        
        node -> val = node -> next -> val;
        node -> next = node -> next -> next;
        
        return head;
    }
};