反转链表 II（leetcode）

题目链接：. - 力扣（LeetCode）

分享两道解题思路：

第一个：

将left~right之间的节点翻转，首先left前的节点的next置为空，right的指针置为空，

然后拼接

p1指的是left前面的一个 p1->next = NULL

p2指的是left

p3指的是right p3->next = NULL;

p4指的是right后的一个

注意：当p1为空时，即left为1，返回值就不是头结点，而是翻转后的第一个节点

复制代码

 struct ListNode* reserve( struct ListNode* phead )
 {
    if( !phead || !phead->next )return phead;
    struct ListNode* p1 = NULL;
    struct ListNode* p2 = phead;
    struct ListNode* p3 = phead->next;
    while( p2 )
    {
        p2->next = p1;
        p1 = p2;
        p2 = p3;
        if( p3 )p3 = p3->next;
    }
    return p1;
}
struct ListNode* reverseBetween(struct ListNode* head, int left, int right) {
    if( left == right )
    return head;
    int i = 0;
    struct ListNode* phead = head;
	struct ListNode* p1 = NULL;
	struct ListNode* p2 = NULL;
	struct ListNode* p3 = NULL;
	struct ListNode* p4 = NULL;
	for( i = 1 ; head ; i++)
	{
		if( i == left - 1 )
		{
			p1 = head;
			struct ListNode* temp ;
			temp = head->next;
			head = temp;
		}
		else if( i == left )
		{
			p2 = head;
			head = head->next;
		}
		else if( i == right  )
		{
			p3 = head;
			p4 = head->next;
			struct ListNode* temp ;
			temp = head->next;
			head->next = NULL;
			head = temp;
		}
		else
		{
			head = head->next;
		}
	}
	struct ListNode* temp  = reserve(p2);
	if( p1 )
	p1->next = temp;
	p2->next = p4;
	if( !p1 )return temp;
	return phead;
}

思路二：

栈处理：将left~right之间的节点存入栈中，然后再将栈顶数据倒出

要有哨兵节点：原因：当left为1时，头结点改变（会影响返回值），为了不让头结点改变，增加头结点

复制代码

typedef struct Stack
{
	struct ListNode** arr;
	int size;
	int top;
}ST;
void StackInit(ST* obj)
{
	obj->size = 0;
	obj->top = -1;
	obj->arr = (struct ListNode**)malloc(sizeof(struct ListNode*)*10000);
}
void StackPush( ST* obj,struct ListNode* temp )
{
	obj->arr[++obj->top] = temp;
	obj->size++;
}
void StackPop( ST* obj )
{
	obj->size--;
	obj->top--;
}
bool StackIsEmpty(ST* obj )
{
	return obj->size==0;
}
struct ListNode* StackTop( ST* obj )
{
	return obj->arr[obj->top];
}
struct ListNode* reverseBetween(struct ListNode* head, int left, int right) {
    ST obj;
	StackInit(&obj);
	struct ListNode* phead = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode* temp = phead;
	phead->next = head;
	struct ListNode* p = head;
	int i = 1;
	while( p )
	{
		if( i == left )
		{
			while( i <= right  )
			{
				i++;
				StackPush(&obj,p);
				p = p->next;
			}
			while( !StackIsEmpty(&obj) )
			{
				phead->next = StackTop(&obj);
				phead = phead->next;
				StackPop(&obj);
			}
			phead->next = p;
		}
		else
		{
			i++;
			phead->next = p;
			phead = phead->next;
			p = p->next;
		}
	}
	return temp->next;
}