1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
|
// 1237. Find Positive Integer Solution for a Given Equation
#include "leetcode.h"
/*
* given a callable function 'f(x, y)' with a hidden formula and a value 'z',
* reverse engineer the formula and return all positive integer pairs 'x' and
* 'y' where 'f(x, y) == z'. you may return the pairs in any order. while the
* exact formula is hidden, the function is monotonically increasing.
*/
int function_id = 0;
int customFunction(int x, int y) {
switch (function_id) {
case 1:
return x + y;
case 2:
return x * y;
}
}
int **findSolution(int (*customFunction)(int, int), int z, int *returnSize,
int **returnColumnSizes) {
int **ans = (int **)malloc(z * sizeof(int *)), k = 0;
int *cols = *returnColumnSizes = (int *)malloc(z * sizeof(int));
for (int i = 1; i <= z; i++)
for (int j = z; j >= 1; j--)
if (customFunction(i, j) == z) {
(ans[k] = (int *)malloc(sizeof(int[cols[k] = 2])))[0] = i;
ans[k++][1] = j;
}
*returnSize = k;
return ans;
}
int main() {
function_id = 1;
int rs1, *rcs1, **fs1 = findSolution(customFunction, 5, &rs1, &rcs1);
for (int i = 0; i < rs1; i++) {
for (int j = 0; j < rcs1[i]; j++)
printf("%d ", fs1[i][j]); // expect: [[1,4],[2,3],[3,2],[4,1]]
printf("\n");
}
function_id = 2;
int rs2, *rcs2, **fs2 = findSolution(customFunction, 5, &rs2, &rcs2);
for (int i = 0; i < rs2; i++) {
for (int j = 0; j < rcs2[i]; j++)
printf("%d ", fs2[i][j]); // expect: [[1,5],[5,1]]
printf("\n");
}
for (int i = 0; i < rs1; i++)
free(fs1[i]);
free(rcs1);
free(fs1);
for (int i = 0; i < rs2; i++)
free(fs2[i]);
free(rcs2);
free(fs2);
}
|