題目

分別實現兩個函式，一個可以刪除單鏈表中倒數第K個節點，另一個可以刪除雙鏈表中倒數第K個節點。

要求

如果連結串列長度為n，則要時間複雜度達到O(N)，額外空間複雜度達到O(1).

思路

一種方法是設定快慢指標，快指標先走k步，此時慢指標從頭開始走，兩者同時開始走，當快指標走到尾時，慢指標走到倒數第K個節點，刪除當前慢指標的節點，（問題是要記錄前後兩個節點）單鏈表和雙鏈表兩種情況。

書中給出的方法是根據k值得變化來進行刪除操作，連結串列從頭開始，每移動一步，k值減1，最後，若k=0：刪除頭節點；若k>0，不刪除；若k<0,當k=0時為倒數第k個節點；之後再次從頭節點開始，k每步自增1，等於0時為倒數k個節點的前一個節點。
在雙鏈表中執行刪除操作時注意prev指標的指向問題即可，不再對雙鏈表進行測試。

以下程式碼根據上面兩個思路均進行了測試。

程式碼

#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;
}