3 个 n^2 的 排序

 1:  /* 因为冒泡是相邻交换, sign 标记着这次不用交换, 下次也不用交换. */
 2:  #include <cstdio>
 3:  
 4:  int main(void) {
 5:      int sign = 1;
 6:      int a[10] = {1, 454, 634,34, 3, 3, 2343, 23435, 34, 42};
 7:  
 8:      for(int i = 0; i < N-1 && sign; ++i) {
 9:          sign = 0;
10:          for(int j = 0; j < N-1-i; ++j) {
11:              if(a[j] > a[j+1]) {
12:                  int temp = a[j];
13:                  a[j] = a[j+1];
14:                  a[j+1] = temp;
15:                  sign = 1;
16:              }
17:          }
18:      }
19:      for(int i = 0; i < 10; ++i) {
20:          printf("%d ", a[i]);
21:      }
22:      printf("\n");
23:  
24:      return 0;
25:  }
26:  

 1:  #include <cstdio>
 2:  
 3:  int main(void) {
 4:      int a[10] = {1, 454, 634,34, 3, 3, 2343, 23435, 34, 42};
 5:  
 6:      for(int i = 0; i < 10-1; ++i) {
 7:          int k = i;
 8:          for(int j = i+1; j < 10; ++j) {
 9:              if(a[j] < a[k]) {   // 选择出最小的
10:                  k = j;
11:              }
12:          }
13:          if(k != i) {
14:              int temp = a[k];
15:              a[k] = a[i];
16:              a[i] = temp;
17:          }
18:      }
19:  
20:      for(int i = 0; i < 10; ++i) {
21:          printf("%d ", a[i]);
22:      }
23:      printf("\n");
24:  
25:      return 0;
26:  }

 1:  #include <cstdio>
 2:  
 3:  int main(void) {
 4:      int a[10] = {1, 454, 634,34, 3, 3, 2343, 23435, 34, 42};
 5:      int i, j;
 6:  
 7:      for(i = 1; i < 10; i++) {
 8:          int key = a[i];
 9:          for(j = i-1; j >= 0; j--) {
10:              if(key < a[j]) {
11:                  a[j+1] = a[j];   // 把牌一直向后推, 只要推完最后一个再插进去就是正确的
12:              }
13:          }
14:          a[j+1] = key;  // j 已经等于 -1 或者 已经越界, 需要加 1
15:      }
16:  
17:      for(int i = 0; i < 10; ++i) {
18:          printf("%d ", a[i]);
19:      }
20:      printf("\n");
21:  
22:      return 0;
23:  }