19. Remove Nth Node From End of List (M)

Remove Nth Node From End of List (M)

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

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题意

给定一个链表，删除倒数第n个结点。

思路

One Pass方法：先将一个指针移动到正数第n个结点处，这时再建一个指针指向头结点，之后同步向后移动这两个指针。当第一个指针到达尾结点时，第二个指针正好在倒数第n个结点处。为了进行删除操作，还要建一个指针保存第二个指针的前结点，要注意删除的正好是头结点这种情况。

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代码实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        int count = 1;
        ListNode last = head;
        while (count != n) {
            last = last.next;
            count++;
        }
        ListNode nth = head;
        ListNode prev = null;	// 只有当nth移动至少一次，prev才会指向具体的结点
        while (last.next != null) {
            last = last.next;
            prev = nth;
            nth = nth.next;
        }
        // 当prev仍为null，说明nth并没有移动，即要删除的就是头结点
        if (prev == null) {
            return nth.next;
        }
        prev.next = nth.next;
        nth.next = null;
        return head;
    }
}