728x90
https://www.hackerrank.com/challenges/breaking-best-and-worst-records/problem
문제는 쉬운 문제인데 함수의 리턴 타입을 고려해서 풀어야하는 문제이다.
int* breakingRecords(int scores_count, int* scores, int* result_count) {
int *breaking=malloc(sizeof(int)*2);
breaking[0]=0; breaking[1]=0;
int min=scores[0], max=scores[0];
for(int i=1;i<scores_count;i++){
if(min>scores[i]) {
breaking[1]++;
min=scores[i];
}
if(max<scores[i])
{
breaking[0]++;
max=scores[i];
}
}
*result_count = 2;
return breaking;
}
△ 작성한 부분
breakingRecords 함수의 return-type은 int* type으로 리턴값으로 전달할 배열인 breaking 의 자료형을 주의해서 코드를 작성해야 한다.
처음에는 int *breaking[2] = {0,0}; 으로 했었는데 오류가 발생했다. 이 때에는 의미가 달라져서 *breaking[0] = 0, *breaking[1] = 0이 된 것 같다. 따라서 breaking 배열을 포인터타입으로 선언하고 malloc으로 동적할당해 배열로 활용한다.
#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* readline();
char** split_string(char*);
int* breakingRecords(int scores_count, int* scores, int* result_count) {
int *breaking=malloc(sizeof(int)*2);
breaking[0]=0; breaking[1]=0;
int min=scores[0], max=scores[0];
for(int i=1;i<scores_count;i++){
if(min>scores[i]) {
breaking[1]++;
min=scores[i];
}
if(max<scores[i])
{
breaking[0]++;
max=scores[i];
}
}
*result_count = 2;
return breaking;
}
int main()
{
FILE* fptr = fopen(getenv("OUTPUT_PATH"), "w");
char* n_endptr;
char* n_str = readline();
int n = strtol(n_str, &n_endptr, 10);
if (n_endptr == n_str || *n_endptr != '\0') { exit(EXIT_FAILURE); }
char** scores_temp = split_string(readline());
int* scores = malloc(n * sizeof(int));
for (int i = 0; i < n; i++) {
char* scores_item_endptr;
char* scores_item_str = *(scores_temp + i);
int scores_item = strtol(scores_item_str, &scores_item_endptr, 10);
if (scores_item_endptr == scores_item_str || *scores_item_endptr != '\0') { exit(EXIT_FAILURE); }
*(scores + i) = scores_item;
}
int scores_count = n;
int result_count;
int* result = breakingRecords(scores_count, scores, &result_count);
for (int i = 0; i < result_count; i++) {
fprintf(fptr, "%d", *(result + i));
if (i != result_count - 1) {
fprintf(fptr, " ");
}
}
fprintf(fptr, "\n");
fclose(fptr);
return 0;
}
char* readline() {
size_t alloc_length = 1024;
size_t data_length = 0;
char* data = malloc(alloc_length);
while (true) {
char* cursor = data + data_length;
char* line = fgets(cursor, alloc_length - data_length, stdin);
if (!line) {
break;
}
data_length += strlen(cursor);
if (data_length < alloc_length - 1 || data[data_length - 1] == '\n') {
break;
}
alloc_length <<= 1;
data = realloc(data, alloc_length);
if (!line) {
break;
}
}
if (data[data_length - 1] == '\n') {
data[data_length - 1] = '\0';
data = realloc(data, data_length);
} else {
data = realloc(data, data_length + 1);
data[data_length] = '\0';
}
return data;
}
char** split_string(char* str) {
char** splits = NULL;
char* token = strtok(str, " ");
int spaces = 0;
while (token) {
splits = realloc(splits, sizeof(char*) * ++spaces);
if (!splits) {
return splits;
}
splits[spaces - 1] = token;
token = strtok(NULL, " ");
}
return splits;
}
SMALL
'Programming > C C++' 카테고리의 다른 글
Birthday Chocolate (0) | 2020.08.30 |
---|---|
[DS] Insert a Node at the Tail of a Linked List (0) | 2020.08.23 |
Print the Elements of a Linked List (0) | 2020.08.16 |
Kangaroo (0) | 2020.08.16 |
Find the Merge point of two joined linked lists (0) | 2020.08.09 |