LeetCode | 19. 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.

其实就是删除链表倒数第N个节点.注意链表无头结点,head就是第一个节点.

//自己的代码,9ms AC/** * 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) {        int tot = 1, i = 1;        ListNode* it = head;        while(it->next)        {            tot++;            it = it->next;        }        it = head;//根据题意,head就是第一个元素        if(n == tot)//删除第一个元素        {            head = head->next;            return head;        }        while(i<tot-n)        {            i++;            it = it->next;        }        if(n == 1)//删除最后一个        {            it->next = NULL;            return head;        }        ListNode* it2 = it->next;        ListNode* it3 = it2->next;        it->next = it3;        it2->next = NULL;        delete(it2);        return head;    }};

//后面提供的代码,用快指针和慢指针,只需遍历一遍.9ms AC/** * 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 == NULL)        return NULL;        ListNode new_head(-1);        new_head.next = head;        ListNode *slow = &new_head, *fast = &new_head;        for (int i = 0; i < n; i++)            fast = fast->next;        while (fast->next)         {            fast = fast->next;            slow = slow->next;        }        ListNode *to_be_deleted = slow->next;   //待删指针为slow->next        slow->next = slow->next->next;        delete to_be_deleted;        return new_head.next;    }};