Leetcode Remove Nth Node From End of List

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.

题意转换一下就是求解单链表的倒数第k个节点，解法是设置两个指针，一个在前一个在后，之间的距离为k，同时向前移动，有点类似滑动窗口，当第一个指针到达链表尾的时候，第二个指针即为倒数第k个节点，具体的见编程之美

ListNode *removeNthFromEnd(ListNode *head, int n){
    if(head == NULL) return NULL;
    ListNode *q = head,*p = head;
    for(int i = 0 ; i < n - 1; ++ i)
        q = q -> next;
    ListNode *pPre = NULL;
    while(q->next){
        pPre = p;
        p = p->next;
        q = q->next;
    }
    if(pPre == NULL){
        pPre = p->next;
        delete p;
    }else{
        pPre ->next = p ->next;
        delete p;
    }
    return head;
}

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Leetcode Remove Nth Node From End of List

原文：http://www.cnblogs.com/xiongqiangcs/p/3629728.html