根据给定的int型数组,给出全排列
比如:
[1,2] => [[1,2],[2,1]
[1,2,3] =>[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
#include <stdio.h>
#include <stdlib.h>
int *concat(int* list, int length, int x) {
int *result = (int*)malloc((length + 1) * sizeof(int));
for (int i = length; i > 0; i--) {
result[i] = list[i - 1];
}
result[0] = x;
return result;
}
int getFactorial(int n) {
int total = 1;
for (int i = 1; i <= n; i++) {
total *= i;
}
return total;
}
int** getPermutation(int * list ,int n) {
int** result;
int total = getFactorial(n);
if (n == 2) {
result = (int**)malloc(total * sizeof(int*));
for (int i = 0; i < total; i++) {
result[i] = (int*)malloc(n * sizeof(int));
}
result[0][0] = list[0];
result[0][1] = list[1];
result[1][0] = list[1];
result[1][1] = list[0];
return result;
}else{
result = (int**)malloc(total * sizeof(int*));
for (int i = 0; i < n; i++) {
int* tempList = (int*)malloc((n - 1) * sizeof(int));
int curIdx = 0;
for (int j = 0; j < n; j++) {
if (i == j) continue;
tempList[curIdx] = list[j];
curIdx++;
}
int** tempResult = getPermutation(tempList, n - 1);
int tempResultLength = getFactorial(n - 1);
for (int k = 0; k < tempResultLength; k++) {
result[tempResultLength * i + k] = concat(tempResult[k], n - 1, list[i]);
}
free(tempList);
free(tempResult);
}
return result;
}
}
int main() {
int n =4;
int* p = (int*)malloc(n * sizeof(int));
for (int i = 0; i < n; i++) {
p[i] = i + 1;
}
int **result = getPermutation(p,n);
for (int i = 0; i < getFactorial(n); i++) {
for (int j = 0; j < n; j++) {
printf("%d,", result[i][j]);
}
printf("\n");
}
free(p);
free(result);
return 0;
}