c语言字符串习题

反转字符串

#include 
#include <string.h>
#include <assert.h>
void reverse_string(char *start, char *end)
{
    assert(start != NULL && end != NULL);
    while(start < end)
    {
        int tmp = *start;
        *start = *end;
        *end = tmp;
        start++;
        end--;
    }
}
int main()
{
    char arr[] = "i love you";
    reverse_string(arr, arr+strlen(arr)-1);
    printf("%s", arr);
}

字符串左旋k位

根据上题的启发，可以利用三次反转

#include 
#include <string.h>
#include <assert.h>
void reverse_string(char *start, char *end)
{
    assert(start != NULL && end != NULL);
    while(start < end)
    {
        int tmp = *start;
        *start = *end;
        *end = tmp;
        start++;
        end--;
    }
}

char * left_remove(char * arr, int k)
{
    int len = strlen(arr)-1;
    char *ret = arr;
    assert(arr != NULL);
    reverse_string(arr, arr+len);
    reverse_string(arr, arr+len-k);
    reverse_string(arr+len-k+1, arr+len);
    return ret;
}
int main()
{
    char arr[] = " i love you";
    int kpoi = 2;
    printf("%s", left_remove(arr, kpoi));
}

查找一个字符串中第一个只出现1(K)次的字符

此题重要的是建立hash思想，空间换时间

#include 
#include 
#include 

int find_first(char * arr, int * Count, int k)
{
    int i = 0;
    assert(arr != NULL);
    for(i=0; i<strlen(arr); i++)
    {
        Count[arr[i]]++;
    }
    for(i=0; i<strlen(arr); i++)
    {
        if(Count[arr[i]] == k)
        {
            return arr[i];
        }
    }
    return -1;
}
int main()
{   
    char arr[] = "i really really like you";
    int arrCount[255] = {0};
    int k = 1;
    int ret = 0;
    if((ret = find_first(arr, arrCount, k)) != -1)
    {
        printf("%c", ret);
    }
    else
    {
        printf("未找到");
    }
}

输出数字三角形

1
22
333
4444
55555

等图案

#include 
#include 
#include 
void print_first(int n)
{
    int i = 0;
    int j = 0;
    for(i=1; i<=n; i++)
    {
        for(j=0; jprintf("%d", i);
        }
        printf("\n");
    }
}
void print_second(int n)
{
    int i = 0;
    int j = 0;
    for(i=1; i<=n; i++)
    {
        for(j=0; jprintf(" ");
        }
        printf("%d\n", i);
    }
}
void print_third(int n)
{
    int i = 0;
    int j = 0;
    for(i=1; i<=n; i++)
    {
        for(j=0; jprintf(" ");
        }
        for(j=0; jprintf("%d", i);
        printf("\n");
    }
}
int main()
{   
    int n = 4;
    print_first(n);
    print_second(n);
    print_third(n);
}

一个正整数有可能可以被表示为 m(m>=2) 个连续正整数之和

15 = 1 + 2 + 3 + 4 + 5
15 = 4 + 5 + 6
15 = 7 + 8
编写一个程序，输入一个正整数，然后找出符合这种要求的所有
连续正整数序列，若不存在这种序列，则打印None。

#include 
#include 
#include 
void print_sequence(int n)
{
    int i = 0;
    int k = 0;
    for(i=1; i<=n/2; i++)
    {
        int sum = 0;
        int tmp = i;
        while(sum < n)
        {
            sum += tmp;
            if(sum == n)
            {
                for(k=i; k<=tmp; k++)
                {
                    printf("%d", k);
                }
                printf("\n");
            }
            tmp++;
        }
    }
}
int main()
{   
    int n = 15;
    print_sequence(n);
}

调整数组使奇数全部都位于偶数前面

两个指针进行移动，交换。在指针移动中要考虑特殊情况（全为奇数）

#include 
#include 
#include 

void remove_odd(int arr[], int size)
{
    int left = 0;
    int right = size -1;
    int tmp = 0;
    assert(arr != NULL);
    while(left < right)
    {
        while(left1) == 1))
        {
            left++;
        }
        while(left1) != 1))
        {
            right--;
        }
        tmp = arr[left];
        arr[left] = arr[right];
        arr[right] = tmp;
    }
}
int main()
{   
    int i = 0;
    int arr[] = {8,3,9,5,2,7,4,6};
//  int arr[] = {1,3,5,7,9};
    int size = sizeof(arr)/sizeof(arr[0]);
    remove_odd(arr, size);
    for(i=0; iprintf("%d", arr[i]);
    }
}

杨氏矩阵

复杂度小于O（n）的查找，每次进行一步查找可以筛选掉一行或一列。

#include 
#include 
#include 
#define ROW 3
#define COL 3
int search_num(int arr[ROW][COL], int key, int *x, int *y)
{
    int xTmp = *x;
    int yTmp = *y;
    assert(x!= NULL && y!=NULL && arr!= NULL);
    while(xTmp>=0 && yTmpif(key == arr[xTmp][yTmp]){
            *x = xTmp;
            *y = yTmp;
            return 1;
        }
        else if(key < arr[xTmp][yTmp])
        {
            xTmp--;
        }
        else if(key > arr[xTmp][yTmp])
        {
            yTmp++;
        }
    }
    return -1;
}
int main()
{   
    int key = 1;
    int arr[ROW][COL] = {1,2,3,2,3,4,3,4,5};
    int ret = 0;
    int x = ROW-1, y = 0;
    if(ret = search_num(arr, key, &x, &y) != -1)
    {
        printf("%d的坐标是%d %d", key, x, y);
    }
    else
    {
        printf("未找到");
    }
}

判断一个字符串是否为回文字符串

#include 
#include 
#include 

int judge_string(char arr[])
{
    int left = 0;
    int right = strlen(arr) -1;
    while(left < right)
    {
        if(arr[left] != arr[right])
        {
            return 0;
        }
        left++;
        right--;
    }
    return 1;
}
int main()
{   
    char arr[] = "abccba";
    int ret = judge_string(arr);
    if(ret)
    {
        printf("是回文字符串");
    }
    else
    {
        printf("不是回文字符串");
    }
}

判断一个数是否为回文数

#include 
#include 
#include 

int judge_num(int num)
{
    int tmp = 0;
    int newNum = 0;
    int copyNum = num;
    while(num>0)
    {
        tmp = num%10;
        num /= 10;
        newNum *= 10;
        newNum += tmp;
    }

    if(newNum == copyNum)
    {
        return 1;
    }
    return 0;
}
int main()
{   
    int num = 12344321;
    int ret = judge_num(num);
    if(ret == 1)
    {
        printf("是回文数");
    }
    else
    {
        printf("不是回文数");
    }
}