19. Remove Nth Node From End of List - jiejackyzhang/leetcode-note GitHub Wiki

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.

Linked List类题目。

这里的要求是one pass，因此我们用两个指针first和second，两者相差n+1个节点，这样当first到达尾节点的next（null）时，second则是指着倒数n+1个节点，这样只需将second的next指向second.next.next节点，即可将倒数第n个节点删除。

注意点是要删除的可能是头节点，因此我们可以首先创建一个dummy节点，指向头节点，first和second初始都指向dummy。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode first = dummy;
        ListNode second = dummy;
        // Advances first pointer so that the gap between first and second is n nodes apart
        for (int i = 1; i <= n + 1 && first != null; i++) {
            first = first.next;
        }
        // Move first to the end, maintaining the gap
        while (first != null) {
            first = first.next;
            second = second.next;
        }
        second.next = second.next.next;
        return dummy.next;
    }
}
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