在单链表和双链表中删除倒数第K个节点

题目

分别实现两个函数，一个可以删除单链表中倒数第K个节点，另一个可以删除双链表中倒数第K个节点。

要求

如果链表长度为n，则要时间复杂度达到O(N)，额外空间复杂度达到O(1).

思路

一种方法是设置快慢指针，快指针先走k步，此时慢指针从头开始走，两者同时开始走，当快指针走到尾时，慢指针走到倒数第K个节点，删除当前慢指针的节点，（问题是要记录前后两个节点）单链表和双链表两种情况。

书中给出的方法是根据k值得变化来进行删除操作，链表从头开始，每移动一步，k值减1，最后，若k=0：删除头节点；若k>0，不删除；若k<0,当k=0时为倒数第k个节点；之后再次从头节点开始，k每步自增1，等于0时为倒数k个节点的前一个节点。
在双链表中执行删除操作时注意prev指针的指向问题即可，不再对双链表进行测试。

以下代码根据上面两个思路均进行了测试。

代码

#include<iostream>
using namespace std;

struct singleList {
	int value;
	singleList* next;
	singleList(int val)
	{
		value = val;
	}
};

struct doubleList {
	int value;
	doubleList* next;
	doubleList* prev;
	doubleList(int val)
	{
		value = val;
	}
};

void delLastKSingle1(singleList* head, int k)
{
	if (head == NULL || k <= 0)
		return;
	singleList* rawHead = new singleList(-1);
	rawHead->next = head;
	singleList* fast = head;
	singleList* slow = head;
	for (int i = 0; i < k; i++)
	{
		fast = fast->next;
	}
	singleList* prev = rawHead;
	while (fast != NULL)
	{
		fast = fast->next;
		slow = slow->next;
		prev = prev->next;
	}
	singleList* pNode = slow;
	prev->next = slow->next;
	delete pNode;
}

singleList* delLastKSingle(singleList* head, int k)
{
	if (head == NULL || k <= 0)
		return head;
	singleList* cur = head;
	while (cur != NULL)
	{
		k--;
		cur = cur->next;
	}
	if (k == 0)
		head = head->next;
	if (k < 0)
	{
		cur = head;
		while ((k++) != 0)
			cur = cur->next;
		cur->next = cur->next->next;
	}
	return head;
}

singleList* createSingle(int* in, int len)
{
	singleList* head = new singleList(in[0]);
	singleList* p = head;
	for (int i = 1; i < len; i++)
	{
		singleList* node = new singleList(in[i]);
		p->next = node;
		p = p->next;
	}
	p->next = NULL;
	return head;
}

void printS(singleList* head)
{
	cout << "print node of list:" << endl;
	while (head != NULL)
	{
		cout << head->value << " ";
		head = head->next;
	}
	cout << endl;
}

int main()
{
	int input[] = { 1, 2, 3, 4, 5 };
	int len = 5;
	singleList* headS = createSingle(input, len);
	int k = 2;
	//delLastKSingle1(headS, k);
	delLastKSingle1(headS, k);
	printS(headS);
	getchar();
	return 0;
}