基本问题
如何删除单链表中的倒数第n个节点？
常规解法
先遍历一遍单链表，计算出单链表的长度，然后，从单链表头部删除指定的节点。
代码实现
/**
*
* Description: 删除单链表倒数第n个节点，常规解法.
*
* @param head
* @param n
* @return ListNode
*/
public static ListNode removeNthFromEnd(ListNode head, int n) {
if (head == null )
return null ;
//get length of list
ListNode p = head;
int len = 0;
while (p != null ) {
len++;
p = p. next ;
}
//if remove first node
int fromStart = len - n + 1;
if (fromStart == 1)
return head. next ;
//remove non-first node
p = head;
int i = 0;
while (p != null ) {
i++;
if (i == fromStart - 1) {
p. next = p. next . next ;
}
p = p. next ;
}
return head;
}
一次遍历法
使用快慢指针。快指针比慢指针提前n个单元。当快指针到达单链表尾部时，慢指针指向待删除节点的前节点。
代码实现
/**
*
* Description: 删除单链表倒数第n个节点，快慢指针法.
*
* @param head
* @param n
* @return ListNode
*/
public static ListNode removeNthFromEnd(ListNode head, int n) {
if (head == null )
return null ;
ListNode fast = head;
ListNode slow = head;
for ( int i = 0; i < n; i++) {
fast = fast. next ;
}
//if remove the first node
if (fast == null ) {
head = head. next ;
return head;
}
while (fast. next != null ) {
fast = fast. next ;
slow = slow. next ;
}
slow. next = slow. next . next ;
return head;
}
