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A slice function at codility
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#include <stdio.h> | |
#include <stdlib.h> | |
#include <assert.h> | |
#include <time.h> | |
#define ARRAY_SIZE 10000 | |
#define MAX 100 | |
int solution1(int A[], int N) { | |
int total = 0; | |
for (int i = 0; i < N; i++) { | |
int tmp = 0; | |
for (int j = i; j < N; j++) { | |
tmp += A[j]; | |
if (tmp == 0) { | |
total++; | |
} | |
} | |
} | |
return total; | |
} | |
int solution2(int A[], int N) { | |
int total = 0; | |
int* p = (int *)malloc(sizeof(int) * N); | |
for (int i = 0; i < N; i++) { | |
p[i] = 0; | |
} | |
for (int i = 0; i < N; i++) { | |
int tmp = 0; | |
int last = i; | |
if (p[i] > 0) { | |
int k = i; | |
while (p[k] > 0 && k < N) { | |
total++; | |
k = p[k]; | |
} | |
continue; | |
} | |
for (int j = i; j < N; j++) { | |
tmp += A[j]; | |
if (tmp == 0) { | |
total++; | |
if (0 < p[j + 1]) { | |
int k = j + 1; | |
while (p[k] > 0 && k < N) { | |
total++; | |
k = p[k]; | |
} | |
break; | |
} | |
p[last] = j + 1; | |
last = j + 1; | |
} | |
} | |
} | |
free(p); | |
return total; | |
} | |
int main(int argc, const char* argv[]) { | |
time_t t; | |
srand((unsigned) time(&t)); | |
int* ns = malloc(sizeof(int) * ARRAY_SIZE); | |
for (int i = 0; i < ARRAY_SIZE; i++) { | |
ns[i] = (rand() % MAX * 2) - MAX; | |
} | |
double t1 = 0; | |
double t2 = 0; | |
int res1 = -1; | |
int res2 = -1; | |
int start = 0; | |
for (int i = 0; i < 100; i++) { | |
// Simple algorithm | |
start = clock(); | |
res1 = solution1(ns, ARRAY_SIZE); | |
t1 += (double)(clock() - start) * 1000 / CLOCKS_PER_SEC; | |
// Faster algorithm | |
start = clock(); | |
res2 = solution2(ns, ARRAY_SIZE); | |
t2 += (double)(clock() - start) * 1000 / CLOCKS_PER_SEC; | |
// Make sure the answer is the same | |
assert(res1 == res2); | |
} | |
printf("solution1: (%.3fms)\n", t1 / 100); | |
printf("solution2: (%.3fms)\n", t2 / 100); | |
return 0; | |
} |
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The result.
The solution2 is much faster than the first one.