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C 2015年笔试题【保】

1、编写一个完整的程序,使之能完成以下功能:从键盘中输入若干个整数,用链表储存这些输入的数,并要求存储的顺序与输入的顺序相反。

分析:链表建立【头插法】

代码:

#include <stdio.h>
#include <stdlib.h>

//定义单链表
typedef struct slist{
    int data;
    struct slist *next;
};

void main()
{
    struct slist *head,*temp; //head为头结点
    head = (struct slist *)malloc(sizeof(struct slist));
    head ->next = NULL;
    int i,s;

    printf("输入元素:\n");
    scanf("%d",&s);
    while(s != 9999)
    {
        temp = (struct slist *)malloc(sizeof(struct slist));
        temp ->data = s;
        //头插法,建立逆序链表
        temp ->next = head ->next;
        head ->next = temp;
        scanf("%d",&s);
    }

    printf("输出链表:\n");
    temp = head ->next;
    while(temp != NULL)
    {
        printf("%d\t",temp ->data);
        temp = temp ->next;
    }
}

2、编写一个函数,把整数序列分成两个部分,使得左边部分都不大于右边部分,不需要排序。 ( 考察的是快速排序的部分)

 函数代码:

//链表划分
int partition(int arr[],int low,int high)
{
    int pos=low;
    int i=low, j=high;
    int temp;

    while(i!=j)
    {
        while(arr[j]>=arr[pos] && i<j)j--;
        while(arr[i]<=arr[pos] && i<j)i++;
        if(i<j)
        {
            temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    temp = arr[i];
    arr[i] = arr[pos];
    arr[pos] = temp;

    return i;
}

完整代码:

#include <stdio.h>
#include <stdlib.h>
const int m = 7;

//链表划分
int partition(int arr[],int low,int high)
{
    int pos=low;
    int i=low, j=high;
    int temp;

    while(i!=j)
    {
        while(arr[j]>=arr[pos] && i<j)j--;
        while(arr[i]<=arr[pos] && i<j)i++;
        if(i<j)
        {
            temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    temp = arr[i];
    arr[i] = arr[pos];
    arr[pos] = temp;

    return i;
}

//快速排序递归式
void quicksort(int a[],int low,int high)
{
    if(low < high)
    {
        int pivotpos = partition(a,low,high);
        quicksort(a,low,pivotpos - 1);
        quicksort(a,pivotpos+1,high);
    }
}

void main()
{
    int a[] = {1,4,7,8,5,2,3};
    int low = 0, high = m - 1,i;
    quicksort(a,low,high);

    printf("排序后:\n");
    for(i = 0; i < m; i++)
    {
        printf("%d\t",a[i]);
    }
}
  

3、有两个整数数组AB,它们分别有mn个整数。并且都是按非递减序列,现将B数组插入A数组中,使得A数组中各元素不大于B数组中各元素,非递减序列。

实例:B[2,3,6,9] 数组插入A[1,4,7] 数组,顺序排序

思路一:将B插入A后,再顺序排序

代码:

#include <stdio.h>
#include <stdlib.h>

void main()
{
    int n = 3,m = 4;
    int a[7] = {1,4,7},b[] = {2,3,6,9},i = 0,j;
    j = n;
    //将B插入到A后
    while(i < m)
    {
        a[j++] = b[i++];
    }
    //将A进行排序【冒泡】
    for(i = 0;i < m+n-1;i++)
    {
        for(j =0;j < m+n-1-i;j++)
        {
            int tmp;
            if(a[j] > a[j+1])
            {
                tmp = a[j+1];
                a[j+1] = a[j];
                a[j] = tmp;
            }
        }
    }
    printf("将B插入A中,顺序排序:\n");
    for(i = 0;i < m+n;i++)
    {
        printf("%d\t",a[i]);
    }
}

思路二:使用附加数组C

代码:

#include <stdio.h>
#include <stdlib.h>
const int n = 3;
const int m = 4;
void main()
{
    int a[] = {1,4,7},b[] = {2,3,6,9},c[7] = {0},i = 0,j = 0,k = 0;
    do
    {
        if(a[i] > b[j])
        {
            c[k++] = b[j++];
        }
        else
            c[k++] = a[i++];
        if(i >= n)
            c[k++] = a[j++];
        if(j >= m)
            c[k++] = b[i++];
    }while((i >= n && j >= m) == 0);

    printf("将B插入A中,顺序排序:\n");
    for(i = 0;i < m+n;i++)
    {
        printf("%d\t",c[i]);
    }
}

4、两个递增有序整数数列链表LaLb,将他们合并后,变成一个新的链表,要求该链表递减排序。结点node由整型data节点指针next构成

代码:

#include <stdio.h>
#include <stdlib.h>
typedef struct slist { int data; struct slist *next; }; //合并【头插法逆序】 void unionslist(struct slist *la,struct slist *lb) { struct slist *p=la->next; struct slist *q=lb->next; struct slist *temp; la->next=0; while(p&&q) { if(p->data <= q->data) { temp = p->next; p->next = la->next; la->next = p; p=temp; } else { temp = q->next; q->next = la->next; la->next = q; q=temp; } } //若La比Lb短,La遍历完,Lb依序插入 if(q) //若q不为空 p=q; while(p) { temp = p->next; p->next = la->next; la->next = p; p=temp; } } //打印输出 void input(struct slist *head) { struct slist *p = head ->next;//p为工作指针 while(p != NULL) { printf("%d\t",p ->data); p = p ->next; } } int main() { //构建链表La和Lb struct slist *La,*Lb,*p;// La为La链表的头结点,Lb为Lb链表的头结点,p为尾指针 La = (struct slist*)malloc(sizeof(struct slist)); Lb = (struct slist*)malloc(sizeof(struct slist)); La ->next = NULL; Lb ->next = NULL; int a[] = {1,4,7,8},b[] = {2,3,5,6,9},i,j; p = La; for(i = 0; i < 4; i++) { //尾插法顺序 struct slist *node = (struct slist*)malloc(sizeof(struct slist)); node ->data = a[i]; node ->next = p ->next; p ->next = node; p = node; } p = Lb; for(i = 0; i < 5; i++) { //尾插法顺序 struct slist *node = (struct slist*)malloc(sizeof(struct slist)); node ->data = b[i]; node ->next = p ->next; p ->next = node; p = node; } //输出 printf("La:"); input(La); printf("\nLb:"); input(Lb); //合并 printf("\n合并后:"); unionslist(La,Lb); input(La); return 0; }

5、编写一个函数,删除链表中的最小值。结点node由整型data节点指针next构成

代码:

#include <stdio.h>
#include <stdlib.h>

typedef struct slist
{
    int data;
    struct slist *next;
};
//删除最小值
void delmin(struct slist *la)
{
    struct slist *p = la ->next,*pre = la;  //p为工作指针,pre为p的前驱,min记录最小值,premin是min的前驱
    struct slist *min = p,*minpre = la;
    while(p)
    {
        if(p ->data < min ->data)
        {
            min = p;
            minpre = pre;
        }
        pre = p;
        p = p ->next;
    }
    minpre ->next = min ->next;
    free(min);
}
//打印输出
void input(struct slist *head)
{
    struct slist *p = head ->next;//p为工作指针
    while(p != NULL)
    {
        printf("%d\t",p ->data);
        p  = p ->next;
    }
}
int main()
{
    //构建链表La
    struct slist *La,*p;// La为La链表的头结点,p为尾指针
    La = (struct slist*)malloc(sizeof(struct slist));
    La ->next = NULL;
    int a[] = {3,6,9,8,1,5,2},i;
    p = La;
    for(i = 0; i < 7; i++)
    {
        //尾插法顺序
        struct slist *node = (struct slist*)malloc(sizeof(struct slist));
        node ->data = a[i];
        node ->next = p ->next;
        p ->next = node;
        p = node;
    }
    //输出
    printf("La:");
    input(La);
    //删除最小值
    delmin(La);
    //删除最小值后输出
    printf("\n删除最小值后La:");
    input(La);

    return 0;
}

6、编写函数判断小括号是否匹配。

分析:由于限定了只有小括号,故不需要栈来实现

代码:

#include <stdio.h>
#include <stdlib.h>

int ismarry(const char *str)
{
    int top=-1;
    char c;

    while(*str)
    {
        if(*str == '(')++top;
        if(*str == ')')
        {
            if(top == -1)return -1; //若此时指针指向“)”,top = -1,这说明前面的已匹配成功
            top--;
        }
        str++;
    }
    if(top == -1)return 0;
    return -1;
}

int main()
{
    char str[] = "((())))";
    if(ismarry(str) == 0)
        printf("匹配成功!");
    else
        printf("匹配不成功!");
    return 0;
}

7、【】对多个字符串进行字典排序

对一个字符串进行字典排序:

 代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void Sort(char parr[],int n)
{
    int i,j;
    char *str1, *str2;
    //冒泡排序
    for(i=0; i<n-1; i++)
    {
        for(j=i+1; j<=n-1; j++)
        {
            str1=parr[i];
            str2=parr[j];
            while(((*str1) && (*str2) && (*str1==*str2)) == 1)
            {
                str1++;
                str2++;
            }
            if(*str1-*str2>0)
            {
                char *temp = parr[i];
                parr[i] = parr[j];
                parr[j] = temp;
            }
        }
    }
}
int main()
{
    char ch[]  = "helloworld!";
    printf("排序前:%s",ch);  //字符串输出
    Sort(ch,strlen(ch));
    printf("\n排序后:%s",ch);  //字符串输出
    return 0;
}

对多个字符串按字典顺序排序

代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//对数组内前n个字符串进行字典排序
int main()
{
    char *str[] = {"hello","world","shao","hang"};  //指针数组
    int i,j,n;
    char *temp;

    scanf("%d",&n);
    for(i =0; i < n-1; i++)
    {
        for(j = i+1; j <n; j++)
        {
            //strcmp(str1,str2),比较函数,若str1 > str2,则返回正数
            if(strcmp(str[i],str[j]) > 0)
            {
                temp = str[j];
                str[j] = str[i];
                str[i] = temp;
            }
        }
    }
    for(i = 0;i < n;i++)
    {
        printf("%s\t",str[i]);
    }
    return 0;
}

8、【不懂】编写一个函数,使之能完成以下功能:利用递归方法找出一个数组中的最大值和最小值

要求递归调用函数的格式如下:MinMaxValue(arr,n,&max,&min),其中arr是给定的数组,n是数组的个数,max、min分别是最大值和最小值

代码:

void MinMaxValue(int arr[], int n, int *max, int *min)
{
	if(n==1)
	{
		*max = arr[0];
		*min = arr[0];
	}
	else
	{
		int _max=arr[0], _min=arr[0];
		MinMaxValue(arr+1, n-1, max, min);
		if(*max<_max)*max=_max;
		if(*min>_min)*min=_min;
	}
}

9、有两字符数组st,求ts中出现第一次的开始位置,如果没有则输出“No”,有则输出开始位置

 代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int start(char s[],char t[])
{
    int s_length = strlen(s);
    int i;
    char *str1, *str2;
    for(i=0; i<s_length; i++)
    {
        str1 = s+i;
        str2 = t;
        while(*str1 && *str2 && *str1==*str2)str1++,str2++; //str1与str2相同,
        if(*str1-*str2 == *str1)  //str2为空,判断str2在sre1中
        {
            printf("%d",i);
            return 0;
        }

    }
    printf("NO!\n");
    return -1;
}
int main()
{
    char a[] = "helloworld",b[] = "o";
    start(a,b);
    return 0;
}
posted @ 2020-03-27 17:11  Pam/sh  阅读(...)  评论(...编辑  收藏