Remove Nth Node From End of List leetcode

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.

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我使用两个指针,两个指针之间的距离保持相差n
较慢的指针指向的就是需要删除的结点
ListNode* removeNthFromEnd(ListNode* head, int n) {
    ListNode dummy(0);
    dummy.next = head;
    ListNode *slow = &dummy, *fast = &dummy;
    while (n--)
        fast = fast->next;
    while (fast->next != nullptr)
    {
        slow = slow->next;
        fast = fast->next;
    }
    ListNode *temp = slow->next;
    slow->next = slow->next->next;
    delete temp;
    return dummy.next;
}
原文地址:https://www.cnblogs.com/sdlwlxf/p/5100177.html