链表-删除、合并 - HelloWorld开发者社区

链表

删除链表中的某个节点或某一段区间

leetcode.203

链接https://leetcode.cn/problems/...
解题方法：链表中删除一个节点的常规方法就是找到这个节点的前驱节点，将前驱节点的next指针指向当前节点的后继节点

leetcode解题代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* removeElements(ListNode* head, int val) {
      auto dummy = new ListNode(-1);
      dummy->next = head;

      for (auto p = dummy; p; p = p->next){
          auto q = p->next;
          while (q && q->val == val) q = q->next;
          p->next = q;
      }
      return dummy->next;
  }
};

leetcode.19

链接https://leetcode.cn/problems/...
解题方法：与上一题类似，需要先求出链表总长度，再找到当前节点的前驱节点
注意：当前节点的前驱节点为倒数第n+1个点，即为正数第len-n个点
需要移动len-n-1次移动到前驱节点

leetcode解题代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* removeNthFromEnd(ListNode* head, int n) {
      auto dummy = new ListNode(-1);
      dummy->next = head;

      int len = 0;
      for (auto p = dummy; p; p = p->next) len ++;

      auto p = dummy;
      for (int i = 0; i < len - n - 1; i ++) p = p->next;
      p->next = p->next->next;
      return dummy->next; 
  }
};

leetcode.237

链接https://leetcode.cn/problems/...
解题方法：本题不是常规意义上的删除节点，无法像上一道题一样找到当前节点的前驱节点
本题的方法比较取巧，即先把后继节点的值赋值给当前节点，然后再删除后继节点

leetcode解题代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  void deleteNode(ListNode* node) {
      node->val = node->next->val;
      node->next = node->next->next;
  }
};

leetcode.2095

链接https://leetcode.cn/problems/...
解题方法：快慢指针
快指针每次向前移动两位，慢指针每次向前移动一位
当快指针走到链表结尾，满指针刚好指向链表中间
通过虚拟头节点让快慢指针都回退一位，满指针则刚好指向中间节点的前驱节点

leetcode解题代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* deleteMiddle(ListNode* head) {
      auto dummy = new ListNode(-1);
      dummy->next = head;
      auto fast = dummy, slow = dummy;
      while (fast && fast->next && fast->next->next){
          fast = fast->next->next;
          slow = slow->next;
      }
      slow->next = slow->next->next;
      return dummy->next;
  }
};

leetcode.83

链接https://leetcode.cn/problems/...
解题方法：双指针
cur指针指向无重复元素链表的最后一个节点
p指针遍历链表，如果p指针元素和cur指针元素不相等则将cur指针指向p，并更新cur指针
最后cur指针是无重复元素链表的最后一个节点，记得将其指向空

leetcode解题代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* deleteDuplicates(ListNode* head) {
      if (!head) return head;
      auto cur = head;
      for (auto p = head->next; p; p = p->next)
          if (p->val != cur->val)
              cur = cur->next = p;
      cur->next = nullptr;
      return head;
  }
};

leetcode.82

链接https://leetcode.cn/problems/...
解题方法：双指针
p指针维护新链表的最后一位，q指针遍历去找无重复元素的区间
如果区间长度=1则说明没有重复元素，保留
如果区间长度>1则说明有重复元素，删除整个区间

leetcode解题代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* deleteDuplicates(ListNode* head) {
      auto dummy = new ListNode(-1);
      dummy->next = head;
      auto p = dummy;
      while (p->next){
          auto q = p->next;
          while (q && q->val == p->next->val) q = q->next;
          if (q == p->next->next) p = p->next;
          else p->next = q;
      }
      return dummy->next;
  }
};

合并两个有序链表

leetcode.21

链接https://leetcode.cn/problems/...
解题方法：归并排序
维护一个当前节点指针
如果list1的值小于list2，指针指向list1并更新当前节点指针和list1的指针
如果list2的值小于list1，指针指向list2并更新当前节点指针和list2的指针
如果两个链表长度不同，当一个链表遍历完成，另一个链表没有遍历完成，当前节点指针指向没有遍历完成的链表

leetcode解题代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
      auto dummy = new ListNode(-1);
      auto cur = dummy;
      while (list1 && list2){
          if (list1->val < list2->val){
              cur = cur->next = list1;
              list1 = list1->next;
          }
          else{
              cur = cur->next = list2;
              list2 = list2->next;
          }
      }
      if (list1) cur->next = list1;
      if (list2) cur->next = list2;
      return dummy->next;
  }
};

leetcode.23

链接https://leetcode.cn/problems/...
解题方法：归并排序与上一题类似
如何从k个链表中找到最小的值呢？维护一个堆
在c++中维护一个堆用priority_queue，默认是大根堆，所以还需要重载一下比较函数

leetcode解题代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  struct Cmp{
      bool operator() (ListNode* a, ListNode* b){
          return a->val > b->val;
      }
  };

  ListNode* mergeKLists(vector<ListNode*>& lists) {
      priority_queue<ListNode*, vector<ListNode*>, Cmp> heap;
      auto dummy = new ListNode(-1);
      auto cur = dummy;
      for (auto c: lists) 
          if (c) heap.push(c);

      while (heap.size()){
          auto t = heap.top();
          heap.pop();

          cur = cur->next = t;
          if (t->next) heap.push(t->next);
      }
      return dummy->next;
  }
};

解题参考：https://www.acwing.com/