链表操作与经典问题解析

1. 常用操作和常用技巧

1.1 常用操作

创建一个新节点

#include <iostream>
using namespace std;

struct ListNode {
    int val;          // 节点的值
    ListNode* next;   // 指向下一个节点的指针

    // 构造函数初始化节点
    ListNode(int value) : val(value), next(nullptr) {}
};

尾插

void tailInsert(ListNode*& head, int value) {
    ListNode* newNode = new ListNode(value); // 创建新节点
    if (head == nullptr) { // 如果链表为空,直接让新节点作为头节点
        head = newNode;
        return;
    }
    ListNode* temp = head; // 临时指针遍历到链表末尾
    while (temp->next != nullptr) {
        temp = temp->next;
    }
    temp->next = newNode; // 将新节点连接到链表尾部
}

头插

void headInsert(ListNode*& head, int value) {
    ListNode* newNode = new ListNode(value); // 创建新节点
    newNode->next = head; // 新节点指向当前的头节点
    head = newNode;       // 更新头节点为新节点
}

1.2 常用技巧

(1)引入虚拟“头”节点

反转链表

Node* reverseList(Node* head) {
    Node* prev = nullptr;
    Node* cur = head;
    while (cur) {
        Node* nextNode = cur->next;
        cur->next = prev;
        prev = cur;
        cur = nextNode;
    }
    return prev;
}

(2)快慢双指针

判断是否是带环链表

bool hasCycle(Node* head) {
    Node* slow = head;
    Node* fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
        if (slow == fast) return true;
    }
    return false;
}

找到链表中倒数第n个节点

Node* removeNthFromEnd(Node* head, int n) {
    Node* dummy = new Node(0);
    dummy->next = head;
    Node* slow = dummy;
    Node* fast = dummy;
    for (int i = 0; i <= n; ++i) fast = fast->next;
    while (fast) {
        slow = slow->next;
        fast = fast->next;
    }
    slow->next = slow->next->next;
    return dummy->next;
}

2. 两数相加(中等)

模拟两数相加的过程即可

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* newnode = new ListNode; //申请一个新链表节点
        ListNode* pcur = newnode;
        int t = 0; // 记录进位

        while(l1 || l2 || t)
        {
            if(l1 != nullptr) // 加上第一个链表的值
            {
                t += l1->val;
                l1 = l1->next;
            }
            if(l2) //加上第二个链表
            {
                t += l2->val;
                l2 = l2->next;
            }
            pcur->next = new ListNode(t % 10); // 个位
            t /= 10; //得到进位
            pcur = pcur->next;
        }
        pcur = newnode->next;
        delete newnode;

        return pcur;
    }
};

2. 两数相加 - 力扣(LeetCode)

3. 两两交换链表中的节点(中等)

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if(head == nullptr || head->next == nullptr) return head;

        ListNode* newnode = new ListNode(0, head);
        
        ListNode* prev = newnode, *cur = head, *next = cur->next, *nnext = next->next;

        while(cur && next)
        {
            //交换节点
            prev->next = next;
            cur->next = nnext;
            next->next = cur;

            //修改节点
            prev = cur, cur = nnext;
            if(cur)  next = cur->next;
            if(next)  nnext = next->next;
        }
        return newnode->next;
    }
};

24. 两两交换链表中的节点 - 力扣(LeetCode)

4. 重排链表(中等)

下面的思路是将先反转后一半链表,然后前半后半链表接替相加即可。

class Solution {
public:
    void reorderList(ListNode* head) {
        if(head == nullptr || head->next == nullptr || head->next->next == nullptr) return; 

        //1. 快慢指针找到中间节点
        ListNode* fast = head, *slow = head;
        while(fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        }

        //2. 反转后半部分链表 头插法
        ListNode* head2 = new ListNode(0);
        ListNode* cur = slow->next, *next = cur->next;
        slow->next = nullptr;
        while(cur)
        {
            next = cur->next;
            cur->next = head2->next;
            head2->next = cur;
            cur = next;
        }

        //3. 重新排列
        ListNode* ret = new ListNode(0);
        ListNode* cur1 = head, *cur2 = head2->next;
        ListNode* prev = ret;
        while(cur1)
        {
            //先放第一个链表
            prev->next = cur1;
            cur1 = cur1->next;
            prev = prev->next;

            //再放第二个链表
            if(cur2)
            {
                prev->next = cur2;
                cur2 = cur2->next;
                prev = prev->next;
            }
        }
    }
};

143. 重排链表 - 力扣(LeetCode)

5. 合并k个升序链表(困难)

 

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {

        //1. 先计算要翻转多少组链表
        ListNode* ccur = head;
        int n = 0;
        while(ccur)
        {
            ccur = ccur->next;
            n++;
        }
         n /= k;

        //2. 重复n次:长度为k的链表的逆序
        ListNode* newnode = new ListNode(0);
        ListNode* prev = newnode, *tmp = head;
        ListNode *cur = head, *next = cur->next;

        for(int i = 0; i < n; ++i)
        {
            tmp = cur;
            for(int j = 0; j < k; ++j)
            {
                next = cur->next;

                cur->next = prev->next;
                prev->next = cur;

                cur = next;
            }
            prev = tmp;
        }
        prev->next = cur;
        cur = newnode->next;
        delete newnode;

        return cur;
    }
};

25. K 个一组翻转链表 - 力扣(LeetCode)

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