(C) 프로그래머스 BCT - Chap.07

 

가까운 1 찾기

#include <stdio.h>

int solution(int arr[], size_t len, int idx) {
    for (int i = 0; i < len; i++)
        if(arr[i] && i >= idx)
            return i;
    return -1;
}

 

길이에 따른 연산

#include <stdio.h>

int sum_all(int num[], int len)
{
    int res = 0;
    
    for(int idx = 0; idx < len; idx++)
        res += num[idx];
    
    return res;
}

int mul_all(int num[], int len)
{
    int res = 1;
    
    for(int idx = 0; idx < len; idx++)
        res *= num[idx];
    
    return res;
}

int solution(int num[], size_t len) {
    if(len > 10)
        return sum_all(num, len);
    return mul_all(num, len);
}

 

대문자로 바꾸기

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

#define TRS(c) ((c) > 96 && (c) < 123 ? -32 : 0)

char* solution(const char* str) {
    char* res = (char*)malloc((strlen(str) + 1) * sizeof(char));
    
    for(int idx = 0; idx < strlen(str); idx++)
        res[idx] = str[idx] + TRS(str[idx]);
    res[strlen(str)] = '\0';
    
    return res;
}

 

문자 개수 세기

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

int* solution(const char* str) {
    int* res = (int*)calloc(52, sizeof(int));
    
    while(*str)
    {
        if(*str >= 'A' && *str <= 'Z')
            res[*str - 'A']++;
        else
            res[*str - 'A' - 6]++;
        str++;
    }
        
    return res;
}

 

문자열로 변환

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

char* solution(int n) {
    char* res = (char*)malloc(6 * sizeof(char));
    int rdx = 0;
    int lst[5] = {0, };
    int ldx = 0;
    
    while(n)
    {
        lst[ldx++] = n % 10;
        n /= 10;
    }
    
    for(int idx = ldx - 1; idx >= 0; idx--)
        res[rdx++] = lst[idx] + 48;
    res[ldx] = NULL;
    
    return res;
}

 

배열 비교하기

#include <stdio.h>

#define COMP(x, y) ((x) > (y) ? 1 : (x) < (y) ? -1 : (x) == (y) ? 0 : 0)

int sum(int arr[], int len)
{
    int res = 0;
    
    for(int idx = 0; idx < len; idx++)
        res += arr[idx];
    
    return res;
}

int solution(int arr1[], size_t arr1_len, int arr2[], size_t arr2_len) {
    if(arr1_len - arr2_len)
        return COMP(arr1_len, arr2_len);
    return COMP(sum(arr1, arr1_len), sum(arr2, arr2_len));
}

 

세로 읽기

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

char* solution(const char* str, int rows, int cols) {
    char* res = (char*)malloc((strlen(str) + 1) * sizeof(char));
    int spc = 0;
    
    for(int idx = cols - 1; idx < strlen(str); idx += rows)
        res[spc++] = str[idx];
    res[spc] = NULL;
    
    return res;
}

 

소문자로 바꾸기

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

#define LOW(n) ((n) < 91 ? (n + 32) : (n))

char* solution(const char* str) {
    char* res = (char*)malloc((strlen(str) + 1) * sizeof(char));
    
    for(int idx = 0; idx < strlen(str); idx++)
        res[idx] = LOW(str[idx]);
    res[strlen(str)] = NULL;
    
    return res;
}

 

수 조작하기 1

#include <stdio.h>

#define CONTROL(x) ((x == 'w') ? 1 : (x == 's') ? -1 : (x == 'd') ? 10 : (x == 'a') ? -10 : 0)

int solution(int n, const char* control) {
    for(int idx = 0; idx < strlen(control); idx++)
        n += CONTROL(control[idx]);
    return n;
}

 

수 조작하기 2

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

char* solution(int log[], size_t len) {
    char* res = (char*)malloc((len) * sizeof(char));
    int spc = 0;
    
    for (int idx = 1; idx < len; idx++)
    {
        int diff = log[idx] - log[idx - 1];
        switch(diff)
        {
            case -10: res[spc++] = 'a'; break;
            case -1: res[spc++] = 's'; break;
            case 1: res[spc++] = 'w'; break;
            case 10: res[spc++] = 'd'; break;
        }
    }
    res[spc] = NULL;
    
    return res;
}

 

정수 부분

#include <stdio.h>

int solution(double flo) {
    return (int)flo;
}

 

정수 찾기

#include <stdio.h>

int solution(int num[], size_t len, int n) {
    for(int idx = 0; idx < len; idx++)
        if(num[idx] == n)
            return 1;
    return 0;
}

 

첫 번째로 나오는 음수

#include <stdio.h>

int solution(int num[], size_t len) {
    for(int idx = 0; idx < len; idx++)
        if(num[idx] < 0)
            return idx;
    return -1;
}

 

n 번째 원소부터

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

int* solution(int num[], size_t len, int n) {
    int* res = (int*)malloc((len - (n - 1)) * sizeof(int));
    int spc = 0;
    
    for(int idx = n - 1; idx < len; idx++)
        res[spc++] = num[idx];
    
    return res;
}

 

n 번째 원소까지

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

int* solution(int num[], size_t len, int n) {
    int* res = (int*)malloc(n * sizeof(int));
    
    for(int idx = 0; idx < n; idx++)
        res[idx] = num[idx];
    
    return res;
}

 

qr code

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

#define QR(q, r, c) ((c) % (q) == (r))

char* solution(int q, int r, const char* code) {
    char* res = (char*)malloc((strlen(code) + 1) * sizeof(char));
    int rdx = 0;
    
    for(int idx = 0; idx < strlen(code); idx++)
        if(QR(q, r, idx))
            res[rdx++] = code[idx];
    res[rdx] = NULL;
    
    return res;
}

 

https://github.com/JenSeop/Programmers-C/tree/main/Chap.07