链表OJ题

简介: 链表OJ题


1. 链表的回文结构

因为单链表不能从后往前找节点,所以我们先找到中间节点,然后逆置,最后进行比较。

#include <stdio.h>
#include <stdbool.h>
typedef struct ListNode
{
  int val;
  struct ListNode* next;
}ListNode;
struct ListNode* middleNode(struct ListNode* head)
{
  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)
{
  if (NULL == head)
  {
    return head;
  }
  ListNode* n1, * n2, * n3;
  n1 = NULL, n2 = head, n3 = head->next;
  while (n2)
  {
    n2->next = n1;
    n1 = n2;
    n2 = n3;
    if (n3)
    {
      n3 = n3->next;
    }
  }
  return n1;
}
bool chkPalindrome(ListNode* A)
{
  ListNode* mid = middleNode(A);
  ListNode* rmid = reverseList(mid);
  while (A && rmid)
  {
    if (A->val != rmid->val)
    {
      return false;
    }
    A = A->next;
    rmid = rmid->next;
  }
  return true;
}

2. 相交链表

#include <stdio.h>
#include <stdlib.h>
struct ListNode
{
  int val;
  struct ListNode* next;
};
struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB)
{
  struct ListNode* curA = headA;
  struct ListNode* curB = headB;
  int lenA = 0;
  while (curA->next)
  {
    ++lenA;
    curA = curA->next;
  }
  
  int lenB = 0;
  while (curB->next)
  {
    ++lenB;
    curB = curB->next;
  }
  //不相交
  if (curA != curB)
  {
    return NULL;
  }
  int gap = abs(lenA - lenB);
  //因为我们不知道A长还是B长,所以我们要用假设法,先假设A长,如果不对,再调换一下就行
  struct ListNode* longList = headA;
  struct ListNode* shortList = headB;
  if (lenB > lenA)
  {
    longList = headB;
    shortList = headA;
  }
  //让长的先走gap步
  while (gap--)
  {
    longList = longList->next;
  }
  //再同时走,找交点
  while (longList != shortList)
  {
    longList = longList->next;
    shortList = shortList->next;
  }
  return longList;
}

3. 链表中倒数第k个结点

思路2:

#include <stdio.h>
struct ListNode
{
  int val;
  struct ListNoe* next;
};
struct ListNode* FindKthToTail(struct ListNode* pListHead, int k)
{
  struct ListNode* slow = pListHead, * fast = pListHead;
  //fast先走k步
  while (k--)
  {
    //k还没有减到0,链表已经空了,说明k大于链表的长度
    if (NULL == fast)
    {
      return NULL;
    }
    fast = fast->next;
  }
  //slow和fast同时走,fast走到空,slow就是倒数第k个
  while (fast)
  {
    slow = slow->next;
    fast = fast->next;
  }
  return slow;
}

4. 环形链表

#include <stdbool.h>
struct ListNode
{
  int val;
  struct ListNode* next;
};
 
bool hasCycle(struct ListNode* head)
{
  struct ListNode* slow = head, * fast = head;
  while (fast && fast->next)
  {
    slow = slow->next;
    fast = fast->next->next;
    if (slow == fast)
    {
      return true;
    }
  }
  return false;
}

5. 环形链表 II

找入环点:

法一:

#include <stdio.h>
struct ListNode
{
  int val;
  struct ListNode* next;
};
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;
}

法二:

#include <stdio.h>
#include <stdlib.h>
struct ListNode
{
  int val;
  struct ListNode* next;
};
struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB)
{
  struct ListNode* curA = headA;
  struct ListNode* curB = headB;
  int lenA = 0;
  while (curA->next)
  {
    ++lenA;
    curA = curA->next;
  }
  int lenB = 0;
  while (curB->next)
  {
    ++lenB;
    curB = curB->next;
  }
  //不相交
  if (curA != curB)
  {
    return NULL;
  }
  int gap = abs(lenA - lenB);
  struct ListNode* longList = headA;
  struct ListNode* shortList = headB;
  if (lenB > lenA)
  {
    longList = headB;
    shortList = headA;
  }
  //让长的先走gap步
  while (gap--)
  {
    longList = longList->next;
  }
  //再同时走,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)
    {
      struct ListNode* meet = slow;
      struct ListNode* headx = meet->next;
      meet->next = NULL;
      return getIntersectionNode(head, headx);
    }
  }
  return NULL;
}

6. 随机链表的复制

写代码的时候建议一边画细图,一边写:

#include <stdio.h>
#include <stdlib.h>
struct Node
{
    int val;
    struct Node *next;
    struct Node *random;
};
struct Node* copyRandomList(struct Node* head)
{
    struct Node* cur = head;
    //拷贝节点插入原节点的后面
    while (cur)
    {
        struct Node* copy = (struct Node*)malloc(sizeof(struct Node));
        copy->val = cur->val;
        //插入
        copy->next = cur->next;
        cur->next = copy;
        //迭代
        cur = cur->next->next;
    }
    //控制拷贝节点的random
    cur = head;
    while (cur)
    {
        struct Node* copy = cur->next;
        
        if (NULL == cur->random)
        {
            copy->random = NULL;
        }
        else
        {
            copy->random = cur->random->next;
        }
        //迭代
        cur = cur->next->next;
    }
    //把copy节点解下来,链接成新链表
    struct Node* copyhead = NULL, * tail = NULL;
    cur = head;
    while (cur)
    {
        struct Node* copy = cur->next;
        struct Node* next = copy->next;
        //尾插
        if (NULL == tail)
        {
            copyhead = tail = copy;
        }
        else
        {
            tail->next = copy;
            tail = tail->next;
        }
        cur->next = next;
        cur = next;
    }
    return copyhead;
}


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