【数据结构】链表的一些面试题

简单不先于复杂，而是在复杂之后。

链表面试题

1. 删除链表中等于给定值 val 的所有结点。OJ链接

//1.常规方法

struct ListNode* removeElements(struct ListNode* head, int val) {
    
    struct ListNode* cur = head, *prev = NULL;
    while(cur)
    {
    
        if(cur->val == val)
        {
    
            //1.头删
            //2.非头删

            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;
}

//2.新思路
//创建一个新链表，把非val的结点尾插到新链表
struct ListNode* removeElements(struct ListNode* head, int val) {
    
   struct ListNode* newhead =NULL, *cur = head, *tail = NULL;
   while(cur)
   {
    
       if(cur->val != val)
       {
    
           if(tail == NULL)
           {
    
               newhead = tail = cur;
           }
           else
           {
    
               tail->next = cur;
               tail = tail->next;
           }
           cur = cur->next;
       }
       else
       {
    
            struct ListNode* del = cur;
            cur = cur->next;
            free(del);
       }
   }

   //最后的节点是val就会出现此问题

   if(tail)
   {
    
       tail->next = NULL;
   }


   return newhead;
}

小技巧：在涉及到链表的题的时候调试不太方便，我们需要快速构建一个链表。

int main()
{
    
    struct ListNode* n1 = (struct ListNode*)malloc(sizeof(struct ListNode));
    assert(n1);
      struct ListNode* n2 = (struct ListNode*)malloc(sizeof(struct ListNode));
    assert(n2);
      struct ListNode* n3 = (struct ListNode*)malloc(sizeof(struct ListNode));
    assert(n3);
      struct ListNode* n4 = (struct ListNode*)malloc(sizeof(struct ListNode));
    assert(n4);
      struct ListNode* n5 = (struct ListNode*)malloc(sizeof(struct ListNode));
    assert(n5);
      struct ListNode* n6 = (struct ListNode*)malloc(sizeof(struct ListNode));
    assert(n6);
      struct ListNode* n7 = (struct ListNode*)malloc(sizeof(struct ListNode));
    assert(n7);

    n1->next = n2;
    n2->next = n3;
    n3->next = n4;
    n4->next = n5;
    n5->next = n6;
    n6->next = n7;
    n7->next = NULL;

    n1->val = 1;
    n2->val = 2;
    n3->val = 6;
    n4->val = 3;
    n5->val = 4;
    n6->val = 5;
    n7->val = 6;

    return 0;
}

带哨兵位的链表：

• 哨兵节点（Sentinel Node）： 这是一个特殊的节点，它不包含实际数据，仅用于简化代码逻辑。哨兵节点通常作为链表的头部存在，这意味着链表中始终有一个节点，即使链表为空。哨兵节点的存在简化了对链表头部的特殊处理，因为无论链表是否为空，我们都可以始终通过哨兵节点来引导链表的访问。
• 优点： 哨兵节点可以简化代码实现，避免在对头部进行操作时需要额外的条件检查。

不带哨兵位的链表：

• 特殊处理头部： 在不使用哨兵节点的情况下，需要特殊处理链表头部，因为在空链表或者插入第一个节点时，需要进行额外的条件检查，以确保正确处理链表头部。
• 优点： 节省了一个节点的空间开销，因为没有用于哨兵的额外节点。

带哨兵位和不带哨兵位的区别：

//利用带哨兵位的头节点的链表可以免去判空的步骤

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)
   {
    
       if(cur->val != val)
       {
    
            tail->next = cur;
            tail = tail->next;
            cur = cur->next;
       }
       else
       {
    
            struct ListNode* del = cur;
            cur = cur->next;
            free(del);
       }
   }

   //最后的节点是val就会出现此问题

   if(tail)
   {
    
       tail->next = NULL;
   }

    head = guard->next;
    free(guard);
    return head;
}

替代之前实现单链表时候传参用的二级指针有两种方式：

• 返回新的链表头指针
• 设计为带哨兵位的链表

(单链表在实际应用中很少带头，OJ题链表基本不带头)

2. 反转一个单链表。OJ链接

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;
}

//思路2：将链表的指针指向反转
struct ListNode* reverseList(struct ListNode* head) {
    
    struct ListNode* prev = NULL;
    struct ListNode* cur = head;
    while(cur)
    {
    
        struct ListNode* next = cur->next;
        cur->next = prev;
        //迭代器
        prev = cur;
        cur = next;
    }
    return prev;
}

3. 给定一个带有头结点 head 的非空单链表，返回链表的中间结点。如果有两个中间结点，则返回第二个中间结点。OJ链接

struct ListNode* middleNode(struct ListNode* head) {
    
    struct ListNode* fast,*slow;
    fast = slow = head;
    while(fast && fast->next)
    {
    
        slow = slow->next;
        fast = fast->next->next;
    }

    return slow;
}

4. 输入一个链表，输出该链表中倒数第k个结点。OJ链接

struct ListNode* FindKthToTail(struct ListNode* pListHead, int k ) {
    
    // write code here
    struct ListNode* slow, *fast;
    slow = fast = pListHead;

    while(k--)
    {
    
       //判断k大于链表长度
        if(!fast)
        {
    
            return NULL;
        }
         fast = fast->next;
    }
    while(fast)
    {
    
        slow = slow->next;
        fast = fast->next;

    }

    return slow;
}

5. 将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有结点组成的。OJ链接

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. 编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大小或等于x的结点之前。OJ链接

class Partition {
    
  public:
    ListNode* partition(ListNode* pHead, int x) {
    
        // write code here
       struct ListNode* lessTail,*lessGuard,*greaterTail,*greaterGuard;
       lessTail = lessGuard = (struct ListNode*)malloc(sizeof(struct ListNode));
       greaterTail = greaterGuard = (struct ListNode*)malloc(sizeof(struct ListNode));
       struct ListNode* cur = pHead;

       greaterGuard->next = NULL;
       lessGuard->next = NULL;

       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。OJ链表

class PalindromeList {
    
  public:
    struct ListNode* middleNode(struct ListNode* head) {
    
        struct ListNode* fast, *slow;
        fast = slow = head;
        while (fast && fast->next) {
    
            slow = slow->next;
            fast = fast->next->next;
        }

        return slow;
    }

    struct ListNode* reverseList(struct ListNode* head) {
    
        struct ListNode* prev = NULL;
        struct ListNode* cur = head;
        while (cur) {
    
            struct ListNode* next = cur->next;
            cur->next = prev;
            //迭代器
            prev = cur;
            cur = next;
        }
        return prev;
    }

    bool chkPalindrome(ListNode* head) {
    
        // write code here
        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;

    }
};

8. 输入两个链表，找出它们的第一个公共结点。OJ链表

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    
    struct ListNode* curA = headA, *curB = headB;

    if(headA == NULL || headB == NULL)
    {
    
        return NULL;
    }

    int lenA = 1;
    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;
    }

    int gap = abs(lenA - lenB);
    while(gap--)
    {
    
        longList = longList->next;
    }

    while(longList != shortList)
    {
    
        longList = longList->next;
        shortList = shortList->next;
    }

    return longList;
}

9. 给定一个链表，判断链表中是否有环。OJ链表

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;
}

【思路】

快慢指针，即慢指针一次走一步，快指针一次走两步，两个指针从链表起始位置开始运行，如果链表带环则一定会在环中相遇，否则快指针率先走到链表的末尾。

【扩展问题】

• 为什么快指针每次走两步，慢指针走一步可以？

假设链表带环，两个指针最后都会进入环，快指针先进环，慢指针后进环。当慢指针刚进环时，可能就和快指针相遇了，最差情况下两个指针之间的距离刚好就是环的长度。

此时，两个指针每移动一次，之间的距离就缩小一步，不会出现每次刚好是套圈的情况，因此：在慢指针走到一圈之前，快指针肯定是可以追上慢指针的，即相遇。

• 快指针一次走3步，走4步，…n步行吗？

  • 快指针一次走 3 步：

    假设slow进环后，fast和slow之间差距N，追赶距离就是N，每追赶一次，之间的距离缩小2步

    如果N是偶数，距离一定会减少到0，也就是两个指针相遇；

    如果N是奇数，距离会减少到1，fast会越过slow指针，fast和slow之间得距离变成了-1,也就是C-1，（C-1是环长度),如果C-1是偶数，再追一圈就可以追上，如果C-1是奇数，永远追不上；

10. 给定一个链表，返回链表开始入环的第一个结点。如果链表无环，则返回 NULL。OJ链接

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;
}

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    
  struct ListNode* curA = headA, *curB = headB;
  
  int lenA = 1;
  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;
  }

  int gap = abs(lenA - lenB);

  while(gap--)
  {
    
      longList = longList->next;
  }

  while(longList != shortList)
  {
    
      longList = longList->next;
      shortList = shortList->next;
  }

  return longList;
}

struct ListNode *detectCycle(struct ListNode *head) {
    
   struct ListNode* slow, *fast;
   slow = fast = head;

   while(fast && fast->next)
   {
    
       slow = slow->next;          
       fast = fast->next->next;

       if(slow == fast)
       {
    
            struct ListNode* meet = slow;
            struct ListNode* next = meet->next;
            meet->next = NULL;
            
            struct ListNode* entry =  getIntersectionNode(head,next);
            meet->next = next;
            return entry;
         
       }
       
   }
   return NULL;
   
}

11. 给定一个链表，每个结点包含一个额外增加的随机指针，该指针可以指向链表中的任何结点或空结点。

​ 要求返回这个链表的深度拷贝。OJ链接

struct Node* copyRandomList(struct Node* head) {
    
	// 1.插入copy节点
    struct Node* cur = head;
    struct Node* copy;
    struct Node* next;
    while(cur)
    {
    
        // 复制链接
        next = cur->next;
        copy = (struct Node*)malloc(sizeof(struct Node));
        copy->val = cur->val;

        cur->next = copy;
        copy->next = next;

        // 迭代往后走
        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;

       //迭代
       cur = copy->next;
   } 

   //3. copy节点要解下来链接在一起，然后恢复原链表
   struct Node* copyHead = NULL, *copyTail = NULL;
   cur = head;
   while(cur)
   {
    
       copy = cur->next;
       next = copy->next;

       //取节点尾插
       if(copyTail == NULL)
       {
    
           copyHead = copyTail = copy;
       }
       else
       {
    
           copyTail->next = copy;
           copyTail = copyTail->next;
       }

       //恢复原链表链接
       cur->next = next;

       //迭代
       cur = next;
   }

   return copyHead;
}

12. 其他。ps：链表的题当前因为难度及知识面等等原因还不适合当前学习，下面有OJ链接。

Leetcode OJ链接 + 牛客 OJ链接