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

 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

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;
}