简单算法随笔（两个整形的交换、二分查找、求n的阶乘 ...）

 

 

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <limits.h>

//两整形的交换
int main(){

    //int a = 3; int b = 5;
    //int tem = 3;
    //a = b;
    //b = tem;
    //printf("%d %d\n", a, b); // 5  3

    //增加条件：不准使用临时变量
    //1:
    //int a = 3; int b = 5;
    //a = a + b;
    //b = a - b;
    //a = a - b;
    //printf("%d %d\n", a, b); // 5 3 此方法有不足，int是4个字节32个比特位最大存储数值是INT_MAX即2147483647(查看需头文件limits.h)，如果超出范围将溢出

    //2:
    int a = 3; int b = 5;
    a = a^b; // 011 ^ 101 = 110
    b = a^b; // 110 ^ 101 = 011
    a = a^b; // 110 ^ 011 = 101
    printf("%d %d\n", a, b); // 5 3 结论：两个数字异或得第三个结果，使用第三个结果异或其中的一个数值得另外一个数字

    return 0;
}

 

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

// 找出只出现过一次的数字

int main(){

    // 1: 此种算法匹配次数较多，可进行优化
    //int arr1[] = {1,2,3,4,5,1,2,3,4};
    //int lenght = sizeof(arr1) / sizeof(arr1[0]);
    //for (int i = 0; i < lenght; i++){  
    //    int count = 0;
    //    for (int j = 0; j < lenght; j++){
    //        if (arr1[i] == arr1[j]){
    //            count++;
    //        };
    //    };
    //    if (count == 1){
    //        printf("%d\n", arr1[i]);
    //        break;
    //    };
    //};

    // 2: 
    int arr1[] = { 1, 2, 3, 4, 5, 1, 2, 3, 4 };
    int lenght = sizeof(arr1) / sizeof(arr1[0]);
    int tem = 0;
    for (int i = 0; i < lenght; i++){
        tem ^= arr1[i]; // 实验得到：a^a = 0; 0^a = a; a^a^b = b;a^b^a = b;
    }
    printf("%d\n", tem);

    return 0;
}

 

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

// 使用递归按顺序打印整形1234
int print(int unm){

    if (unm > 9){
        print(unm / 10);
    };
    printf("%d \n", unm % 10); // 1 2 3 4
}

int main(){
    int unm = 1234;
    print(unm);
    return 0;
}

 

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

// 不创建临时变量求字符串长度

// 1:
//int str(char* one){
//    int count = 0;
//    while(*one != '\0'){
//        count++;
//        one++;
//    }
//    return count;
//}

//2:
int str(char* one){
    if (*one != '\0')
        return 1 + str(one + 1);
    else
        return 0;
}

int main(){

    char arr1[] = "asdfghjkl";
    int lenght = str(arr1);
    printf("%d\n",lenght);
    return 0;
}

 

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

// 求n的阶乘

// 注：如5的阶乘等于 5*4*3*2*1 = 120
//1:
//int func(int input){
//    int unm = 1;
//    for (int i = 1; i <= input; i++){
//        unm *= i; // 1*1*2*3*4*5...
//    }
//    return unm;
//}
//2:
int func(int input){  // 如果input等于5最后一个等于1将返回1,倒数第二个返回4*1给倒数第三个依此类推
    if (input <= 1)
        return 1;
    else
        return input*func(input - 1); // ...5*4*3*2*1  递归一般都会有终止条件、有条件与终止条件越来越接近、最后爆发~
}
int main(){
    int input = 0; int ret = 0;
    printf("请输入一个数字：");
    scanf("%d", &input);
    ret = func(input);
    printf("它的阶乘是 %d\n", ret);
    return 0;
}

 

//求第n个斐波纳契数是什么:
//注斐波纳契数是 1 1 2 3 5 8 13 21 34 55 ...，即前后面的数字是前连个数字的和

//1：
//int func(int n){ // 次方法过于消耗，如果是50则找49,48，49、48有要找依此类推..
//    if (n <= 2)
//        return 1;
//    else
//        return func(n - 1) + func(n-2);
//}

//2:
int func(int n){
    int a = 1; int b = 1; int c = 0;
    while (n>2){
        c = a + b;
        a = b;
        b = c;
        n--;
    };
    return c;
}

int main(){
    int n = 0; int ret = 0;
    scanf("请输入第几个%d",&n);
    ret = func(n);
    printf("第%d个的斐波那契数是：%d",n,ret);
    return 0;
}

 

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

// 算法之二分查找
// 注此算法粗数据有序

int main(){
    int arr[] = {1,2,3,4,5,6,7,8,9,10}; //找出6的索引值

    int lenght = sizeof(arr) / sizeof(arr[0]);
    int left = 0; int right = lenght - 1;

    while (left <= right){
        int index = (left + right) / 2;
        if (arr[index] > 6){
            right = index - 1;
        }
        else if (arr[index] < 6){
            left = index + 1;
        }
        else{
            printf("找到了，它的索引是：%d\n", index);
            break;
        }
    }
    if (left > right){
        printf("找不到\n");
    }

    return 0;
}

 

// 九九乘法表

int main(){

    for (int i = 1; i <= 9;i++){
        for (int j = 1; j <= i;j++){
            printf("%d*%d=%-2d ", i,j,i*j);
        }
        printf("\n");
    }

    return 0;
}

 

// 素数求解(100到200之间)
// 注:素数只能被1和自身整除本案例采用试除法，以后学了其它的再加上去

// 1:
int main(){

    for (int i = 100; i <= 200; i++){
        
        int j = 2;
        for (j = 2; j < 200; j++){
            if (i%j == 0){
                break;
            };
        };
        if (i == j){
            printf("%d ",i);
        }
    }

    return 0;
}

//2(优化)：
int main(){
        for (int i = 101; i <= 200; i+=2)
        {
            int j = 2;
            for (j = 2; j <= sqrt(i); j++)
            {
                if (i%j == 0)
                {
                    break;
                }
            }
            if(j > sqrt(i))
            {
                printf("%d ", i);
            }
        }
        return 0;
}

 

void reverse(int* arr, int len){

    int left = 0; int right = len - 1;
    
    while(left < right){
        int tem = arr[left];
        arr[left] = arr[right];
        arr[right] = tem;
        left++; right--;
    }

}

void print(int* arr, int len){
    for (int i = 0; i < len; i++){
        printf("%d", arr[i]);
    }
}

int main(){
    int arr[10] = {1,2,3,4,5,6,7,8,9,10};
    int len = sizeof(arr) / sizeof(arr[0]);
    reverse(arr,len);
    print(arr, len);
    return 0;
}

 

//交换两个数组的内容
int main(){
    int arr1[] = {1,3,5,7,9};
    int arr2[] = {2,4,6,8,0};
    int len = sizeof(arr1) / sizeof(arr1[0]);
    int tem = 0;
    for (int i = 0; i < len; i++){
        tem = arr1[i];
        arr1[i] = arr2[i];
        arr2[i] = tem;
    }
    printf("%d", arr1[0]);
    return 0;
}

 

//求一个数字的二进制中(补码)有几个1

// 1:
//int countOne(unsigned int a){
//    int count = 0;
//    while (a){
//        if (a % 2 == 1){
//            count++;
//        };
//        a = a / 2;
//    };
//    return count;
//}

// 2:
//int countOne(int a){
//    int count = 0;
//    for (int i = 0; i < 32; i++){
//        if (((a >> i) & 1) == 1){
//            count++;
//        };
//    };
//    return count;
//}

// 3:
// 13二进制为1101；公式： n = n&(n-1)； 结果：1101 n、 1100 n-1、 1100 n、 1011 n-1、 1000 n、 0111 n-1、 0000 n；结论：每次尾数1会消失,且n最终为0
int countOne(int a){
    int count = 0;
    while (a){
        a = a&(a - 1);
        count++;
    }
    return count;
}

int main(){
    int a = 13;
    // 00000000000000000000000000000011
    int count = countOne(a);    
    printf("%d\n", count);
    return 0;
}

 

int func(int m, int n){
    int tem = m^n;
    int count = 0;
    while (tem){
        tem = tem&(tem - 1);
        count++;
    };
    return count;
};
int main(){
    int m = 1999; int n = 2299;
    int count = func(m, n);
    printf("%d\n", count);
    return 0;
}

 

void func(int n){ 
    //奇数
    for (int i = 30; i >= 0; i -= 2){
        printf("%d", (n >> i) & 1);
    }
    printf("\n");

    // 偶数
    for (int i = 31; i >= 0; i -= 2){
        printf("%d", (n >> i) & 1);
    }
    printf("\n");
}

int main(){
    int n = 13;
    func(n);
    return 0;
}

 

int func(unsigned int n){
    if (n > 9){
        return func(n / 10) + n % 10;
    }
    else{
        return n;
    }
}

int main(){
    int n = 1729;
    int unm = func(n);
    printf("%d\n", unm);
    return 0;
}

 

// 判断当前机器是大端存储还是小端存储。**指针类型决定了指针解引用操作符能访问几个字节**；**指针类型决定了指针+1，-1会是几个字节**；

//1:
//int main(){
//    int n = 1;
//    char* p = (char*)&n;
//    if (*p == 1){
//        printf("小端\n");
//    }
//    else{
//        printf("大端\n");
//    }
//    
//    return 0;
//}

//2:
int func(){
    int n = 1;
    return *(char*)&n;
}
int main(){
    int n = func();
    if (n == 1){
        printf("小端\n");
    }
    else{
        printf("大端\n");
    }
    return 0;
}