[数据结构]链表OJ(leetcode)

目录

数据结构之链表OJ：：

                                     1.移除链表元素

                                     2.反转链表

                                     3.链表的中间结点

                                     4.链表中倒数第k个结点

                                     5.合并两个有序链表

                                     6.链表分割

                                     7.链表的回文结构

                                     8.相交链表

                                     9.环形链表

                                    10.环形链表II

                                    11.复制带随机指针的链表

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数据结构之链表OJ：：

1.移除链表元素

删除链表中等于给定值val的所有结点
struct ListNode
{
	int val;
	struct ListNode* next;
};
方法一
struct ListNode* removeElements(struct ListNode* head, int val)
{
	struct ListNode* cur = head, * prev = NULL;
	while (cur != NULL)
	{
		//1.头删
		//2.非头删
		if (cur->val == val)
		{
			if (cur == head)
			{
				head = head->next;
				free(cur);
				cur = head;
			}
			else
			{
				prev->next = cur->next;
				free(cur);
				cur = prev->next;
			}
		}
		else
		{
			prev = cur;
			cur = cur->next;
		}
	}
	return head;
}
不用二级指针的方法:返回头指针
main函数调用方式:plist = removeElements(plist,6);
方法二
struct ListNode* removeElements(struct ListNode* head, int val)
{
	struct ListNode* cur = head;
	struct ListNode* newhead = NULL, * tail = NULL;
	while (cur != NULL)
	{
		if (cur->val != val)
		{
			if (tail == NULL)
			{
				newhead = tail = cur;
			}
			else
			{
				tail->next = cur;
				tail = tail->next;
			}
		}
		else
		{
			struct ListNode* del = cur;
			cur = cur->next;
			free(del);
		}
	}
	if (tail != NULL)
	{
		tail->next = NULL;
	}
	return newhead;
}
方法三:
struct ListNode* removeElements(struct ListNode* head, int val)
{
	struct ListNode* cur = head;
	struct ListNode* guard = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode* tail = guard;
	while (cur != NULL)
	{
		if (cur->val != val)
		{
			tail->next = cur;
			tail = tail->next;
			cur = cur->next;
		}
		else
		{
			struct ListNode* del = cur;
			cur = cur->next;
			free(del);
		}
	}
	最后结点是val 就会出现此问题
	if (tail != NULL)
	{
		tail->next = NULL;
	}
	head = guard->next;
	free(guard);
	return head;
}
替代我们之前实现的二级指针
1.返回新的链表头 2.设计为带哨兵位

2.反转链表

反转链表
方法一:取结点头插到新链表
struct ListNode* reverseList(struct ListNode* head)
{
	struct ListNode* cur = head;
	struct ListNode* newhead = NULL;
	while (cur)
	{
		struct ListNode* next = cur->next;
		cur->next = newhead;
		newhead = cur;
		cur = next;
	}
	return newhead;
}
方法二:翻转链表方向
struct ListNode* reverseList(struct ListNode* head)
{
	struct ListNode* n1, *n2, *n3;
	n1 = NULL;
	n2 = head;
	n3 = NULL;
	while (n2)
	{
		n3 = n2->next;
		n2->next = n1;
		迭代
		n1 = n2;
		n2 = n3;
	}
	return n1;
}

3.链表的中间结点

链表的中间结点——快慢指针
struct ListNode* middleNode(struct ListNode* head)
{
	struct ListNode* slow, * fast;
	slow = fast = head;
	while (fast && fast->next)
	{
		slow = slow->next;
		fast = fast->next->next;
	}
	return slow;
}

4.链表中倒数第k个结点

链表中倒数第k个结点——快慢指针
方法一:
struct ListNode* FindKthToTail(struct ListNode* pListHead, unsigned int k)
{
	struct ListNode* fast, * slow;
	fast = slow = pListHead;
	while (k--) 走k步
	{
		k大于链表长度
		if (fast == NULL)
			return NULL;
		fast = fast->next;
	}
	while (fast)
	{
		slow = slow->next;
		fast = fast->next;
	}
	return slow;
}
方法二:
struct ListNode* FindKthToTail(struct ListNode* pListHead, unsigned int k)
{
	struct ListNode* fast, * slow;
	fast = slow = pListHead;
	while (fast && --k)走k-1步
	{
		k大于链表长度	
		fast = fast->next;
	}
	if (fast == NULL)
		return NULL;
	while (fast->next)
	{
		slow = slow->next;
		fast = fast->next;
	}
	return slow;
}

5.合并两个有序链表 

合并两个有序链表
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2)
{
	struct ListNode* guard = (struct ListNode*)malloc(sizeof(struct ListNode));
	guard->next = NULL;
	struct ListNode* tail = guard;
	struct ListNode* cur1 = list1, * cur2 = list2;
	while (cur1 && cur2)
	{
		if (cur1->val < cur2->val)
		{
			tail->next = cur1;
			cur1 = cur1->next;
		}
		else
		{
			tail->next = cur2;
			cur2 = cur2->next;
		}
		tail = tail->next;
	}
	if (cur1)
		tail->next = cur1;
	if (cur2)
		tail->next = cur2;
	struct ListNode* head = guard->next;
	free(guard);
	return head;
}

6.链表分割 

链表分割
struct ListNode* partition(struct ListNode* pHead, int x)
{
	struct ListNode* lessGuard, * lessTail, * greaterGuard, * greaterTail;
	lessGuard = lessTail = (struct ListNode*)malloc(sizeof(struct ListNode));
	greaterGuard = greaterTail = (struct ListNode*)malloc(sizeof(struct ListNode));
	lessGuard->next = NULL;
	greaterGuard->next = NULL;
	struct ListNode* cur = pHead;
	while (cur)
	{
		if (cur->val < x)
		{
			lessTail->next = cur;
			lessTail = lessTail->next;
		}
		else
		{
			greaterTail->next = cur;
			greaterTail = greaterTail->next;
		}
		cur = cur->next;
	}
	lessTail->next = greaterGuard->next;
	greaterTail->next = NULL;
	pHead = lessGuard->next;
	free(greaterGuard);
	free(lessGuard);
	return pHead;
}

7.链表的回文结构 

链表的回文结构
方法一:找中间结点翻转链表
struct ListNode* middleNode(struct ListNode* head)
{
	struct ListNode* slow, * fast;
	slow = fast = head;
	while (fast && fast->next)
	{
		slow = slow->next;
		fast = fast->next->next;
	}
	return slow;
}
struct ListNode* reverseList(struct ListNode* head)
{
	struct ListNode* n1, * n2, * n3;
	n1 = NULL;
	n2 = head;
	n3 = NULL;
	while (n2)
	{
		n3 = n2->next;
		n2->next = n1;
		n1 = n2;
		n2 = n3;
	}
	return n1;
}
bool chkPalindrome(struct ListNode* head)
{
	struct ListNode* mid = middleNode(head);
	struct ListNode* rmid = reverseList(mid);
	while (head && rmid)
	{
		if (head->val != rmid->val)
			return false;
		head = head->next;
		rmid = rmid->next;
	}
	return true;
}
方法二:整个链表逆置看和链表是否相同 但要注意需要重新拷贝原链表

8.相交链表 

相交链表
判断是否相交判断尾结点的地址是否相同
找交点:求出长度lenA lenB 长的链表先走差距步 第一个相等的就是交点
struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB)
{
	if (headA == NULL || headB == NULL)
	{
		return NULL;
	}
	struct ListNode* curA = headA, * curB = headB;
	int lenA = 1;
	找尾结点
	while (curA->next)
	{
		curA = curA->next;
		++lenA;
	}
	int lenB = 1;
	while (curB->next)
	{
		curB = curB->next;
		++lenB;
	}
	if (curA != curB)
	{
		return NULL;
	}
	struct ListNode* longList = headA, * shortList = headB;
	if (lenA < lenB)
	{
		longList = headB;
		shortList = headA;
	}
	长的链表先走差距步
	int gap = abs(lenA - lenB);
	while (gap--)
	{
		longList = longList->next;
	}
	同时走找交点
	while (longList != shortList)
	{
		longList = longList->next;
		shortList = shortList->next;
	}
	return longList;
}

9.环形链表 

环形链表
快慢指针法:slow进环以后 fast开始追赶slow 
bool hasCycle(struct ListNode* head)
{
	struct ListNode* fast, * slow;
	fast = slow = head;
	while (fast && fast->next)
	{
		slow = slow->next;
		fast = fast->next->next;
		if (slow == fast)
			return true;
	}
	return false;
}
请证明一下，slow走一步 fast一次走两步？
请证明一下，slow走一步 fast一次走三步？是否可以？ 答:不一定
请证明一下，slow走一步 fast一次走X步？是否可以？每次缩小X-1
请证明一下，slow走X步  fast一次走Y步？是否可以？X<Y 每次缩小X-Y
fast走两步时:假设slow进环以后 fast slow之间的差距为N 即追赶距离为N
slow和fast每移动一步 距离缩小1 距离缩小为N N-1 N-2...1 0 距离为0即相遇
fast走三步时:假设slow进环以后 fast slow之间的差距为N 即追赶距离为N
slow每追赶一次 它们之间距离缩小两步 距离变化为N N-2 N-4 N-6...
如果N为偶数则能追上 如果为奇数距离由1变为-1 意味着它们之间的距离变为了C-1(C是环的长度)
如果环减一是偶数再追一圈就能追上 如果环减一为奇数 则永远追不上

10.环形链表II

环形链表II
1.公式证明推导
2.转换成相交问题
fast走的距离 = 2*slow走的距离
假设进环前的长度是L
假设环的长度是C
假设入口点到相遇点距离是X
slow走的距离是L+X
fast走的距离是L+C=X
假设slow进环前 fast在环里面转了N圈 N>=1
2(L+X) = L+X+N*C
(L+X) = N*C
L = N*C-X
L = (N-1)*C+C-X
结论:一个指针A从头开始走 一个指针B从相遇点开始走 它们会在入口点相遇
1.公式证明推导
struct ListNode* detectCycle(struct ListNode* head)
{
	struct ListNode* slow = head, * fast = head;
	while (fast && fast->next)
	{
		slow = slow->next;
		fast = fast->next->next;
		if (slow == fast)
		{
			struct ListNode* meet = slow;
			while (meet != head)
			{
				meet = meet->next;
				head = head->next;
			}
			return meet;
		}
	}
	return NULL;
}
2.转换成相交问题
struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB)
{
	if (headA == NULL || headB == NULL)
	{
		return NULL;
	}
	struct ListNode* curA = headA, * curB = headB;
	int lenA = 1;
	找尾结点
	while (curA->next)
	{
		curA = curA->next;
		++lenA;
	}
	int lenB = 1;
	while (curB->next)
	{
		curB = curB->next;
		++lenB;
	}
	if (curA != curB)
	{
		return NULL;
	}
	struct ListNode* longList = headA, * shortList = headB;
	if (lenA < lenB)
	{
		longList = headB;
		shortList = headA;
	}
	长的链表先走差距步
	int gap = abs(lenA - lenB);
	while (gap--)
	{
		longList = longList->next;
	}
	同时走找交点
	while (longList != shortList)
	{
		longList = longList->next;
		shortList = shortList->next;
	}
	return longList;
}
struct ListNode* detectCycle(struct ListNode* head)
{
	struct ListNode* slow = head, * fast = head;
	while (fast && fast->next)
	{
		slow = slow->next;
		fast = fast->next->next;
		if (slow == fast)
		{
			转换成相交
			struct ListNode* meet = slow;
			struct ListNode* next = meet->next;
			meet->next = NULL;
			struct ListNode* entryNode = getIntersectionNode(head, next);
			恢复环
			meet->next = next;
			return entryNode;
		}
	}
	return NULL;
}

11.复制带随机指针的链表

复制带随机指针的链表
方法一:1.遍历原链表 复制结点 尾插
       2.更新random 找random原链表中第i个 新链表中对应第i个
方法二:1.拷贝原结点 链接到所有原结点的后面
      ps:原结点和拷贝结点建立一个链接关系 找到原结点就可以找到拷贝结点
       2.更新每个拷贝结点的random 
       3.将拷贝结点解下来 链接成新链表
struct Node
{
	int val;
	struct Node* next;
	struct Node* random;
};
struct Node* copyRandomList(struct Node* head)
{
	1.插入copy结点
	struct Node* cur = head;
	struct Node* copy = NULL;
	struct Node* next = NULL;
	while (cur)
	{
		next = cur->next;
		copy = (struct Node*)malloc(sizeof(struct Node));
		copy->val = cur->val;
		cur->next = copy;
		copy->next = next;
		cur = next;
	}
	2.更新copy->random
	cur = head;
	while (cur)
	{
		copy = cur->next;
		if (cur->random == NULL)
			copy->random = NULL;
		else
			copy->random = cur->random->next;
		cur = cur->next->next;
	}
	3.copy结点要解下来链接在一起 恢复原链表
	struct Node* copyHead = NULL, * copyTail = NULL;
	cur = head;
	while (cur)
	{
		copy = cur->next;
		next = copy->next;
		取结点尾插
		if (copyTail == NULL)
		{
			copyHead = copyTail = copy;
		}
		else
		{
			copyTail->next = copy;
			copyTail = copyTail->next;
		}
		恢复原链表链接
		cur->next = next;
		cur = next;
	}
	return copyHead;
}