19. Remove Nth Node From End of List 删除倒数第n各节点

Posted 2020-10-14 hozhangel

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.

 

删除倒数第n个节点，主要有两种做法

法一：先遍历得到链表的长度length，倒数第n个就是正数第length-n+1个（如1 2 3，倒数第三个节点就是3-3+1=1，即第一个节点）

/**
 * 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) {
        ListNode* newlist= new ListNode(0);
        newlist->next = head;  //带头节点的链表
        int length = 0;
        ListNode* first = head;
        while(first != NULL){   //得到链表长度
            length++;
            first = first->next;
        }
        if(n > length) return NULL;
        int n1 = length + 1 - n;//倒数第n个是在正数第n1个节点
        first = newlist; 
        while(n1 - 1 > 0){         //找到目标节点的前一个位置，便于删除目标节点
            n1--;
            first = first->next;
        }
        first->next = first->next->next;  //删除目标节点
        return newlist->next;  //返回不带头节点的链表
    }
};

 

法二：利用两个指针，快指针先走，找到第n个节点，再启动慢指针。

                    此时快慢指针相距n，让两个指针一起走，快指针走到尾节点时，慢指针刚好到倒数第n个节点。

/**
 * 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) {
        if (!head)
            return nullptr;

        ListNode new_head(-1);
        new_head.next = head;

        ListNode *slow = &new_head, *fast = &new_head;

        for (int i = 0; i < n; i++)  //快指针先走到第n个节点
            fast = fast->next;

        while (fast->next)   //快慢指针一起走
        {
            fast = fast->next;
            slow = slow->next;
        }

        ListNode *to_do_deleted = slow->next;   //这两行负责删除操作
        slow->next = slow->next->next;

        delete to_do_deleted;

        return new_head.next;
    }
};