描述：删除单链表的倒数第n个节点
分析：
主要思路是用两个指针遍历单链表，先让右指针前进n个节点，然后再让两个指针同步前进，这样右指针到达表尾null的时候，左指针的下一个即是要删除的节点
for (int i = 1; i <= n + 1; i++) {
right = right.next;
}
for (; right != null; right = right.next) {
left = left.next;
}
为了消除第一个节点的特殊性，用一个头节点指向它
ListNode ahead = new ListNode(0);
ahead.next = head;
删除节点
left.next = left.next.next;
完整实现：
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode ahead = new ListNode(0);
ahead.next = head;
ListNode left = ahead;
ListNode right = ahead;
for (int i = 1; i <= n + 1; i++) {
right = right.next;
}
for (; right != null; right = right.next) {
left = left.next;
}
left.next = left.next.next;
return ahead.next;
}
转载于:https://www.cnblogs.com/hippiebaby/p/5501045.html
