删除单链表倒数第n个节点

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;

}

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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;

}

[/code]

来自为知笔记(Wiz)