C语言刷题 LeetCode 30天挑战 (八)快慢指针法

//Middle of the Linked List

//Given a non-empty, singly linked list with head node head ,

//return a middle node of linked list.

//If there are two middle nodes, return the second middle node

//Example 1:

//Input:1,2,3,4,5]

//0utput:Node 3from this list(Serialization:[3,4,5])

//The returned node has value 3, (The judge's serialization of this node is [3,4,5]).

//Note that we returned a ListNode object ans, such that:

//ans,val =3,ans.next,val =4,ans.next,next.val = 5, and ans.next.next.next = NULL.

//Example 2:

//Input:I[1,2,3,4,5,6]

//Output:Node 4 from this list(serialization:[4,5,6])

//Since the list has two middle nodes with values 3 and 4, we return the second one.

cpp 复制代码
#include <stdio.h>
#include <stdlib.h>

// Definition for singly-linked list.
struct ListNode {
    int val;
    struct ListNode *next;
};

// Function to find the middle node
struct ListNode* middleNode(struct ListNode *head) {
    struct ListNode *slow = head;
    struct ListNode *fast = head;

    // Move slow pointer one step and fast pointer two steps
    while (fast != NULL && fast->next != NULL) {
        slow = slow->next;
        fast = fast->next->next;
    }
    
    // When fast reaches the end, slow is at the middle
    return slow;
}

// Helper function to create a new ListNode
struct ListNode* createNode(int val) {
    struct ListNode* newNode = (struct ListNode*) malloc(sizeof(struct ListNode));
    newNode->val = val;
    newNode->next = NULL;
    return newNode;
}

// Helper function to print the linked list from a given node
void printList(struct ListNode* node) {
    while (node != NULL) {
        printf("%d ", node->val);
        node = node->next;
    }
    printf("\n");
}

int main() {
    // Example 1: [1, 2, 3, 4, 5]
    struct ListNode* head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);
    head->next->next->next = createNode(4);
    head->next->next->next->next = createNode(5);
    
    struct ListNode* middle = middleNode(head);
    printf("Middle node value: %d\n", middle->val);
    printf("List from middle node: ");
    printList(middle);
    
    // Example 2: [1, 2, 3, 4, 5, 6]
    struct ListNode* head2 = createNode(1);
    head2->next = createNode(2);
    head2->next->next = createNode(3);
    head2->next->next->next = createNode(4);
    head2->next->next->next->next = createNode(5);
    head2->next->next->next->next->next = createNode(6);
    
    middle = middleNode(head2);
    printf("Middle node value: %d\n", middle->val);
    printf("List from middle node: ");
    printList(middle);

    return 0;
}
相关推荐
Coder个人博客5 小时前
Linux6.19-ARM64 mm mmu子模块深入分析
大数据·linux·车载系统·系统架构·系统安全·鸿蒙系统
灰子学技术8 小时前
go response.Body.close()导致连接异常处理
开发语言·后端·golang
Doro再努力8 小时前
Vim 快速上手实操手册:从入门到生产环境实战
linux·编辑器·vim
wypywyp8 小时前
8. ubuntu 虚拟机 linux 服务器 TCP/IP 概念辨析
linux·服务器·ubuntu
那个村的李富贵8 小时前
CANN加速下的AIGC“即时翻译”:AI语音克隆与实时变声实战
人工智能·算法·aigc·cann
二十雨辰8 小时前
[python]-AI大模型
开发语言·人工智能·python
Doro再努力8 小时前
【Linux操作系统10】Makefile深度解析:从依赖推导到有效编译
android·linux·运维·服务器·编辑器·vim
senijusene8 小时前
Linux软件编程:IO编程,标准IO(1)
linux·运维·服务器
power 雀儿8 小时前
Scaled Dot-Product Attention 分数计算 C++
算法
Yvonne爱编码8 小时前
JAVA数据结构 DAY6-栈和队列
java·开发语言·数据结构·python