19. Remove Nth Node From End of List

description:

Given a linked list, remove the n-th node from the end of list and return its head.
Note:
Given n will always be valid.
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.

my answer:

感恩

my answer

在链表中，两个有着固定距离的指针一起走的方法要记住，这里是前面那个走到最后一个结点了，后边这个正好走到倒数第n+1个，就是要删掉的那个前面一个

大佬的answer：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(!head->next) return NULL;
        ListNode *pre = head, *cur = head;//结点指针的定义
        for(int i = 0; i < n; ++i) cur = cur->next;
        if(!cur) return head->next;//如果链表长度就是n的话，一个special case
        while(cur->next){
            cur = cur->next;
            pre = pre->next;
        }
        pre->next = pre->next->next;
        return head;
    }
};

relative point get√：

hint :

19. Remove Nth Node From End of List

原文：https://www.cnblogs.com/forPrometheus-jun/p/10668683.html