https://blog.csdn.net/weixin_49303682/article/details/119365319
1 #include
2
3 #define N 9
4
5 void print(int a[])
6 {
7 for(int i = 0; i < N; i++)
8 {
9 printf("%d ", a[i]);
10 }
11 printf("\n");
12 }
13
14 void maopao(int a[])
15 {
16 int i, j;
17 for (i = 0; i < N -1; i++)
18 {
19 for (j = 0; j < N -1 -i; j++)
20 {
21 if (a[j] > a[j + 1])
22 {
23 int temp = a[j];
24 a[j] = a[j+1];
25 a[j+1] = temp;
26 }
27 }
28 }
29 }
30
31 int main()
32 {
33 int a[] = {9,3,1,4,2,7,8,6,5};
34 maopao(a);
35 print(a);
36
37 return 0;
38 }
https://blog.csdn.net/weixin_49303682/article/details/119429850
1 #include
2
3 #define N 9
4
5 void print(int a[])
6 {
7 for(int i = 0; i < N; i++)
8 {
9 printf("%d ", a[i]);
10 }
11 printf("\n");
12 }
13
14 void xuanze(int a[])
15 {
16 int i, j,min,temp;
17 for(i=0;i
https://blog.csdn.net/weixin_49303682/article/details/119430023
1 #include
2
3 #define N 9
4
5 void print(int a[])
6 {
7 for(int i = 0; i < N; i++)
8 {
9 printf("%d ", a[i]);
10 }
11 printf("\n");
12 }
13
14 void xuanze(int a[])
15 {
16 int i,j,tmp;
17 for(i = 1;i < N;i++){
18 tmp = a[i];
19 for(j = i-1; j >= 0;j--)
20 {
21 if(tmp < a[j])
22 {
23 a[j+1] = a[j];
24 }else{
25 break;
26 }
27 }
28 a[j+1] = tmp;
29 }
30 }
31
32 int main()
33 {
34 int a[] = {9,3,1,4,2,7,8,6,5};
35 xuanze(a);
36 print(a);
37
38 return 0;
39 }
https://blog.csdn.net/weixin_49303682/article/details/119364710
1 #include
2
3 #define N 9
4
5 void print(int a[])
6 {
7 for(int i = 0; i < N; i++)
8 {
9 printf("%d ", a[i]);
10 }
11 printf("\n");
12 }
13
14 void shell(int a[])
15 {
16 int i,j,d,tmp;
17 for(d = N/2; d > 0; d /= 2)
18 {
19 for(i = d;i < N;i++)
20 {
21 tmp = a[i];
22 for(j = i-d; j >= 0;j-=d)
23 {
24 if(tmp < a[j])
25 {
26 a[j+d] = a[j];
27 }else{
28 break;
29 }
30 }
31 a[j+d] = tmp;
32 }
33 }
34 }
35
36 int main()
37 {
38 int a[] = {9,3,1,4,2,7,8,6,5};
39 shell(a);
40 print(a);
41
42 return 0;
43 }
https://blog.csdn.net/weixin_49303682/article/details/119364992
1 #include
2
3 #define N 9
4
5 void print(int a[])
6 {
7 for(int i = 0; i < N; i++)
8 {
9 printf("%d ", a[i]);
10 }
11 printf("\n");
12 }
13
14 int kuaisu(int a[],int i,int j)
15 {
16 int tmp;
17 tmp = a[i]; //将a[i]作为基准保存
18 while(i < j)
19 {
20 while(i < j && tmp < a[j])
21 j--;
22 if(i < j)
23 a[i] = a[j];
24 while(i < j && tmp > a[i])
25 i++;
26 if(i < j)
27 a[j] = a[i];
28 }
29 a[i] = tmp;
30 return i;
31 }
32
33 void digui(int a[],int i,int j)
34 {
35 int mid;
36 if(i < j)
37 {
38 mid = kuaisu(a,i,j);
39 digui(a,i,mid-1);
40 digui(a,mid+1,j);
41 }
42 }
43
44 int main()
45 {
46 int a[] = {9,3,1,4,2,7,8,6,5};
47 digui(a,0,N-1);
48 print(a);
49
50 return 0;
51 }
https://blog.csdn.net/qq_63918780/article/details/122527681
1 #include
2 #include
3
4 int func(int *a,int j,int x)
5 {
6 int len = j - 1,i = 0,min;
7 while(i x)
11 {
12 len = min-1;
13 }
14 else if(min < x)
15 {
16 i = min+1;
17 }
18 else
19 {
20 break;
21 }
22 }
23 return min-1;
24 }
25
26 int main()
27 {
28 int j,x,num;
29 int a[] = {1,2,3,4,5,6,7,8,9};
30 printf("输入要查找的数\n");
31 scanf("%d",&x);
32 j = sizeof(a)/sizeof(a[0]);
33 num = func(a,j,x);
34 printf("要查找的数为a[%d]\n",num);
35
36 return 0;
37 }
给定一个包含红色、白色和蓝色,一共 n 个元素的数组,原地对它们进行排序,使得相同颜色的元素相邻,并按照红色、白色、蓝色顺序排列。
此题中,我们使用整数 0、 1 和 2 分别表示红色、白色和蓝色。
https://blog.csdn.net/weixin_38072112/article/details/104375297
1 #include
2 #include
3
4 #define N 6
5
6 void swap(int *a,int *b)
7 {
8 int temp = *a;
9 *a = *b;
10 *b = temp;
11 }
12
13 int main()
14 {
15 int i = 0,start = 0,end = N-1;
16 int a[] = {2,1,1,0,1,0};
17 while(i