我经历的华为一面、二面手撕代码题目（附答案）

一面手撕代码题目

题目描述

给一个链表和一个数，将链表分为两部分，左边部分小于x，右边部分大于或等于x，保证两部分中节点的相对顺序与之前一致。

比如：
Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

我的答案

#include
#define L 6

typedef struct ListNode
{
     
	struct ListNode *next;
	int data;
}Node;

void breakNode(Node* a, Node* b)	// break b from a->b->c
{
     
	a->next = b->next;
}

void insertNode(Node* a, Node* b) // insert a after b
{
     
	if (b && b->next)
	{
     
		Node* temp = b->next;
		b->next = a;
		a->next = temp;
	}
	else
	{
     
		printf_s("cPos or cPos->next is null.");
	}
}

Node* partition(Node *head, int x)
{
     
	Node* temp = head;
	Node* cPos = NULL;
	int found = 0;

	while (temp && temp->next)
	{
     
		if (found == 0 && temp->data < x)
		{
     
			cPos = temp;
			found = 1;
		}
		
		if (found == 1 && temp->next && temp->next->data < x)
		{
     
			Node* ctemp = temp->next;
			breakNode(temp, ctemp);
			insertNode(ctemp, cPos);
			cPos = cPos->next;
		}

		if (temp->next)
		{
     
			temp = temp->next;
		}
		else
		{
     
			break;
		}
		
	}

	return head;
}

void printList(Node *input, int Length)
{
     
	for (int i = 0; i < Length; i++)
	{
     
		printf_s("%d", input->data);
		printf_s(i < Length - 1 ? "->" : "\n");
		input = input->next;
	}
}

void main()
{
     
	int a[L] = {
     1, 4, 3, 2, 5, 2};
	int x = 3;

	Node *input = (Node *)malloc(sizeof(Node));
	input->data = a[0];
	Node *head = input;
	
	for (int i = 1; i < L; i++)
	{
     
		Node *temp = (Node *)malloc(sizeof(Node));
		temp->data = a[i];
		head->next = temp;
		head = head->next;
	}
	head->next = NULL;

	printList(input, L);

	Node *output = partition(input, x);

	printList(output, L);
}

二面手撕代码题目

题目描述

请用随机函数float random()(返回0~1)的随机数，计算出圆周率数值。

我的答案

#include
#include
#include
#include

void main()
{
     
	srand(time(0));
	double x, y, pi;
	long int n, nt = 0;

	printf_s("Input (for example 100000): \n");	// 投针次数
	scanf_s("%ld", &n);

	for (int i = 0; i <= n; i++)
	{
     
		x = rand() / (double)RAND_MAX * 2.0;
		y = rand() / (double)RAND_MAX * 2.0;	// 产生(0, 2) * (0, 2)区间内的随机数		
		if (pow(x - 1.0, 2.0) + pow(y - 1.0, 2.0) <= 1.0) nt++;	//	如果随机点落在圆内
	}
	pi = 4.0 * nt / n;	// pi * r^2 / ((2r)^2) = pi / 4, so * 4.0

	printf_s("pi = %lf\n", pi);
}