单链表——返回中间节点

目录

题目

思路

代码

---

题目

给定一个头结点为 head 的非空单链表，返回链表的中间结点。

如果有两个中间结点，则返回第二个中间结点。

思路

思路1：遍历得到节点数，然后返回中间的节点

思路2：快慢节点，slow每次只走一步，fast每次走两步。

 

代码

struct ListNode* middleNode(struct ListNode* head){
    struct ListNode* cur = head;
	int num = 0;
	while (cur)
	{
		num++;
		cur = cur->next;
	}
	num = (num % 2 == 0) ? (num / 2 + 1) : ((num + 1) / 2);
	cur = head;
	for (int i = 1; i < num; i++)
	{
		cur = cur->next;
	}
	return cur;

}

完整代码

#define _CRT_SECURE_NO_WARNINGS 1
// 给定一个带有头结点 head 的非空单链表，返回链表的中间结点。如果有两个中间结点，则返回第二个中间结点
#include<stdio.h>
#include<stdlib.h>

typedef struct SList
{
	int val;
	struct SList* next;
}SLTNode;

SLTNode* createlist(int* arr, int z)
{
	SLTNode* head = NULL;
	SLTNode* tail = NULL;
	for (int i = 0; i < z; i++)
	{
		SLTNode* newnode = (SLTNode*)malloc(sizeof(struct SList));
		if (newnode == NULL)
		{
			perror("malloc fail");
			exit(-1);
		}
		else
		{
			newnode->val = *(arr + i);
			newnode->next = NULL;
		}
		if (head == NULL)
		{
			head = newnode;
			tail = newnode;
		}
		else
		{
			tail->next = newnode;
			tail = newnode;
		}
	}
	return head;
}
void SLTPrint(SLTNode* head)
{
	SLTNode* cur = head;
	while (cur)
	{
		printf("%d->", cur->val);
		cur = cur->next;
	}
	printf("NULL\n");
}

SLTNode* middleNode(SLTNode* head)
{
	SLTNode* cur = head;
	int num = 0;
	while (cur)
	{
		num++;
		cur = cur->next;
	}
	num = (num % 2 == 0) ? (num / 2 + 1) : ((num + 1) / 2);
	cur = head;
	for (int i = 1; i < num; i++)
	{
		cur = cur->next;
	}
	return cur;
}

void test1()
{
	// 通过数组初始化单链表
	int arr[] = { 1, 2, 3, 4, 5, 6};
	SLTNode* plist = createlist(arr, sizeof(arr) / sizeof(arr[0]));
	SLTPrint(plist);
	// 返回中间节点
	SLTNode* mid = middleNode(plist);
	SLTPrint(mid);
}

int main()
{
	// 返回中间节点
	test1();
	return 0;
}