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:I1,2,3,4,5,6

//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
是隼人3 分钟前
buuctf-pwn bypwn(ret2shellcode)题解(学习过程持续更新)
c语言·学习·安全·pwn入门·ctf入门
Tisfy3 分钟前
LeetCode 3870.统计范围内的逗号:模拟 或 一步计算
数学·算法·leetcode·题解·模拟·遍历
明月_清风12 分钟前
字符串匹配四大经典算法:BF、RK、BM、KMP 到底有什么区别?
后端·算法
疯狂打码的少年12 分钟前
【计算机网络】OSI/RM七层模型(层次结构、各层功能速览)
开发语言·笔记·计算机网络·php
小灰灰搞电子13 分钟前
Rust+Slint 实现ModbusRTU从机调试助手源码分享
开发语言·rust·modbusrtu
程序喵大人14 分钟前
【C++入门】值类别与表达式 - 03 引用绑定:为什么有些参数能接住临时对象
开发语言·c++·引用绑定
子非鱼a24 分钟前
【WEB】[RoarCTF 2019]Easy Java
java·开发语言
软件黑马王子29 分钟前
19.资源加载模块:主要作用和基本原理
开发语言·前端框架·c#
..Dauntless..38 分钟前
【Linux】权限问题——拥有者、所属组和其他用户的协调
linux·运维·服务器
明月_清风40 分钟前
多模式字符串匹配:Trie 与 AC 自动机
后端·算法