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<stdio.h>
#define L 6typedef 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<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<time.h>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 * 4.0printf_s("pi = %lf\n", pi);
}