全排列

https://leetcode-cn.com/problems/permutations/submissions/

  1. int count(int m);
  2. void solution(int* nums, int p, bool* choose, int* temp, int** result, int* m, int numsSize);
  4. int** permute(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
  5.     //初始化标记数组
  6.     bool* choose = malloc(sizeof(bool) * numsSize);
  7.     for(int i = 0; i < numsSize; i++)
  8.         choose[i] = false;
  9.     //存放每一组结果
  10.     int* temp = malloc(sizeof(int) * numsSize);
  11.     //m为全排列行数
  12.     int m = count(numsSize); 
  13.     int** result = malloc(sizeof(int*) * m);
  14.     for(int i = 0; i < m; i++)
  15.         result[i] = malloc(sizeof(int) * numsSize);
  16.     //c为当前行数
  17.     int c = 0;
  18.     int *p = &c;
  19.     solution(nums, numsSize, choose, temp, result, p, numsSize);
  20.     *returnSize = m;
  21.     *returnColumnSizes = (int *)malloc(sizeof(int) * m);
  22.     for (int i = 0; i < m; i++) 
  23.         (*returnColumnSizes)[i] = numsSize;
  24.     return result;
  25. }
  27. int count(int m){
  28.     int count = 1;
  29.     while(m > 0){
  30.         count = count * m;
  31.         m--;
  32.     }
  33.     return count;
  34. }
  36. void solution(int* nums, int p, bool* choose, int* temp, int** result, int* m, int numsSize){
  37.     if(p != 0){
  38.         for(int i = 0; i < numsSize; i++){
  39.             if(*(choose + i))
  40.                 continue;
  41.             choose[i] = true;
  42.             temp[numsSize - p] = nums[i];
  43.             solution(nums, p - 1, choose, temp, result, m, numsSize);
  44.             choose[i] = false;
  45.         }
  46.     }
  47.     else{
  48.         for(int i = 0; i < numsSize; i++)
  49.             result[*m][i] = temp[i];
  50.         (*m)++;
  51.     }
  53. }
  1. int count(int m);
  2. void solution(int* nums, int p, int* temp, int** result, int* m, int numsSize);
  4. int** permute(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
  5.     //存放每一组结果
  6.     int* temp = malloc(sizeof(int) * numsSize);
  7.     //m为全排列行数
  8.     int m = count(numsSize); 
  9.     int** result = malloc(sizeof(int*) * m);
  10.     for(int i = 0; i < m; i++)
  11.         result[i] = malloc(sizeof(int) * numsSize);
  12.     //c为当前行数
  13.     int c = 0;
  14.     int *p = &c;
  15.     solution(nums, numsSize, temp, result, p, numsSize);
  16.     *returnSize = m;
  17.     *returnColumnSizes = (int *)malloc(sizeof(int) * m);
  18.     for (int i = 0; i < m; i++) 
  19.         (*returnColumnSizes)[i] = numsSize;
  20.     return result;
  21. }
  23. int count(int m){
  24.     int count = 1;
  25.     while(m > 0){
  26.         count = count * m;
  27.         m--;
  28.     }
  29.     return count;
  30. }
  32. void swap(int* a, int* b){
  33.     int temp = *a;
  34.     *a = *b;
  35.     *b = temp;
  36. }
  38. void solution(int* nums, int p, int* temp, int** result, int* m, int numsSize){
  39.     if(p != 0){
  40.         for(int i = numsSize - p; i < numsSize; i++){
  41.             temp[numsSize - p] = nums[i];
  42.             swap(&nums[i], &nums[numsSize - p]);
  43.             solution(nums, p - 1, temp, result, m, numsSize);
  44.             swap(&nums[i], &nums[numsSize - p]);
  45.         }
  46.     }
  47.     else{
  48.         for(int i = 0; i < numsSize; i++)
  49.             result[*m][i] = temp[i];
  50.         (*m)++;
  51.     }
  53. }