LeetCode题目

• 203.移除链表元素
• 206.反转链表
• 160.相交链表
• 707.设计链表
• 24.两两交换链表中的节点
• 19.删除链表的倒数第 N 个结点
• 142.环形链表 II

简介

• 链表题目需要注意的点
  • ListNode结构体要会背

  struct ListNode {
        
        
      int val;
      ListNode* next;
  
      ListNode(int _val = 0, ListNode* _next = nullptr)
          : val(_val)
          , next(_next)
      {
        
        
      }
  };
  // 结尾记得加分号
  

  • 先判断head节点是否为nullptr，再进入下一步
  • 有效性判断：要对cur取next，就要先判断cur是否为nullptr，再取next
  • 善用dummy哨兵节点和pre指针
  • 使用pre，cur，next，next2指代已有节点，使用temp指代新建节点
  • 一轮处理完后，更新指针
  • 基本处理模板

  if (!head)
  	return head;
  
  ListNode dummy = ListNode(-1, head);
  ListNode *pre = &dummy, *cur = head, *next;
  while (cur) {
        
        
      next = cur->next;
  
      ...
  
      pre = cur;
      cur = next;
  }
  
  return dummy.next;
  

• 链表题目常用方法与概念
  • 快慢指针，注意有效性判断

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

  • 对撞指针
  • 相交链表，尾部对齐，再逐个检查尾部指针，相同处为相交处
  • 设计链表时，不要单独实现头插和尾插，统一用指定下标插入实现
  • 循环链表，快指针与慢指针重合处，必定为head
  • 有环链表，在快慢指针重合处、head，同时出发，重合点为环的起点

题目讲解

• 以综合题707和易错题24、142为例
  • 707.设计链表

  class MyLinkedList {
        
        
  private:
      struct ListNode {
        
        
          int val;
          ListNode* next;
  
          ListNode(int _val = 0, ListNode* _next = nullptr)
              : val(_val)
              , next(_next)
          {
        
        
          }
      };
  
      ListNode* dummy;
      int64_t size;
  
  public:
      MyLinkedList()
          : dummy(new ListNode(-1, nullptr))
          , size(0)
      {
        
        
      }
  
      int get(int index)
      {
        
        
          if (index > (size - 1) || index < 0)
              return -1;
  
          ListNode *pre = dummy, *cur = dummy->next, *next;
          auto cur_index = index - index;
          while (cur) {
        
        
              next = cur->next;
  
              if (cur_index == index) {
        
        
                  return cur->val;
              }
              ++cur_index;
  
              pre = cur;
              cur = next;
          }
  
          return -1;
      }
  
      // 不要单独实现头插和尾插，统一用指定下标插入实现
      void addAtHead(int val) {
        
         return addAtIndex(0, val); }
  
      void addAtTail(int val) {
        
         return addAtIndex(size, val); }
  
      void addAtIndex(int index, int val)
      {
        
        
          if (index > size || index < 0)
              return;
          if (0 == size) {
        
        
              dummy->next = new ListNode(val);
              ++size;
              return;
          }
  
          ListNode *pre = dummy, *cur = dummy->next, *next;
          auto cur_index = index - index;
          while (cur) {
        
        
              next = cur->next;
  
              if (cur_index == index) {
        
        
                  ListNode* temp = new ListNode(val, cur);
                  pre->next = temp;
                  ++size;
                  break;
              }
              ++cur_index;
  
              pre = cur;
              cur = next;
          }
          if (index == size) {
        
        
              ListNode* temp = new ListNode(val, cur);
              pre->next = temp;
              ++size;
          }
  
          return;
      }
  
      void deleteAtIndex(int index)
      {
        
        
          if (index > (size - 1) || index < 0)
              return;
  
          ListNode *pre = dummy, *cur = dummy->next, *next;
          auto cur_index = index - index;
          while (cur) {
        
        
              next = cur->next;
  
              if (cur_index == index) {
        
        
                  pre->next = next;
                  delete cur;
                  --size;
                  break;
              }
              ++cur_index;
  
              pre = cur;
              cur = next;
          }
  
          return;
      }
  };
  

  • 24.两两交换链表中的节点

  class Solution {
        
        
  public:
      ListNode* swapPairs(ListNode* head)
      {
        
        
          if (!head)
              return head;
          if (!(head->next))
              return head;
  
          ListNode dummy = ListNode(-1, head);
          ListNode *pre = &dummy, *cur = head, *next, *next2;
          while (cur && cur->next) {
        
        
              next = cur->next;
              next2 = next->next;
  
              pre->next = next;
              next->next = cur;
              cur->next = next2;
  
              // 易错点：pre要先更新，并且cur=next2
              pre = cur;
              cur = next2;
          }
  
          return dummy.next;
      }
  };
  

  • 142.环形链表 II

  class Solution {
        
        
  public:
      ListNode* detectCycle(ListNode* head)
      {
        
        
          if (!head)
              return NULL;
          if (!(head->next))
              return NULL;
  
          ListNode dummy = ListNode(-1, head);
          ListNode *slow = head, *fast = head;
          // 易错点：快慢指针有效性判断
          while (slow && fast && fast->next) {
        
        
              slow = slow->next;
              fast = fast->next->next;
  
              if (slow == fast) {
        
        
                  while (slow != head) {
        
        
                      slow = slow->next;
                      head = head->next;
                  }
                  return slow;
              }
          }
  
          return NULL;
      }
  };