求一分治算法

√sdf

有一17个元素的整形数组
寻找最大元素和次大元素

√碧水寒潭

各位帮帮忙啊

═━灆涩━═

先给它排序了,不就行了吗?

X~iao~ping

这是soft_wind提供的递归算法

#include "stdio.h"
#include "conio.h"
void maxmin(int a[],int n,int *max,int *min)
{
int a1[5],a2[5];
int i,j,k=0;
int max1,max2,min1,min2;
if(n==1)
*max=*min=a[0];
else if(n==2)
{
if(a[0]>a[1])
{
*max=a[0];
*min=a[1];
}
else
{
*max=a[1];
*min=a[0];
}
}
else
{
for(i=0;i<n/2;i++)
a1[i]=a[i];
for(j=i;j<n;j++)
a2[k++]=a[j];
maxmin(a1,n/2,&max1,&min1);
maxmin(a2,(n-n/2),&max2,&min2);
if (max1 > max2)
*max = max1;
else
*max = max2;
if (min1 > min2)
*min = min2;
else
*min = min1;
}
}
main()
{
int a[10]={1,12,21,4,57,78,95,43,100,23};
int max,min;
maxmin(a,10,&max,&min);
printf("%d %d",max,min);
getch();
}

生活的成功者

可以用list类，然后再sort();就可以了

青松

1. #include <iostream>
   
2. using namespace std;
   
3. 
   
4. 
   
5. void findMaxAndSubMax(int *pi, int size, int * m, int *sm)
   
6. {
   
7. if(size == 2)
   
8. {
   
9. if(pi[0]>pi[1])
   
10. {
    
11. *m = pi[0];
    
12. *sm = pi[1];
    
13. return;
    
14. }
    
15. *m = pi[1];
    
16. *sm = pi[0];
    
17. return;
    
18. }
    
19. else if(size == 3)
    
20. {
    
21. if(pi[0]>pi[1] && pi[1]>pi[2])
    
22. {
    
23. *m = pi[0];
    
24. *sm = pi[1];
    
25. return;
    
26. }
    
27. else if(pi[0]>pi[2] && pi[2]>pi[1])
    
28. {
    
29. *m = pi[0];
    
30. *sm = pi[2];
    
31. return;
    
32. }
    
33. else if(pi[1]>pi[0] && pi[0]>pi[2])
    
34. {
    
35. *m = pi[1];
    
36. *sm = pi[0];
    
37. return;
    
38. }
    
39. else if(pi[1]>pi[2] && pi[2]>pi[0])
    
40. {
    
41. *m = pi[1];
    
42. *sm = pi[2];
    
43. return;
    
44. }
    
45. else if(pi[2]>pi[0] && pi[0]>pi[1])
    
46. {
    
47. *m = pi[2];
    
48. *sm = pi[0];
    
49. return;
    
50. }
    
51. else if(pi[2]>pi[1] && pi[1]>pi[0])
    
52. {
    
53. *m = pi[2];
    
54. *sm = pi[1];
    
55. return;
    
56. }
    
57. }
    
58. else
    
59. {
    
60. int * m1 = new int;
    
61. int * sm1 = new int;
    
62. int * m2 = new int;
    
63. int * sm2 = new int;
    
64. 
    
65. int firstHalf = size/2;
    
66. int secondHalf = size - firstHalf;
    
67. 
    
68. findMaxAndSubMax(pi, firstHalf, m1, sm1);
    
69. findMaxAndSubMax(pi+firstHalf, secondHalf, m2, sm2);
    
70. 
    
71. int temp1, temp2;
    
72. if(*m1 > *m2)
    
73. {
    
74. *m = *m1;
    
75. temp1 = *m2;
    
76. }
    
77. else
    
78. {
    
79. *m = *m2;
    
80. temp1 = *m1;
    
81. }
    
82. temp2 = (*sm1>*sm2)?*sm1:*sm2;
    
83. *sm = (temp1>temp2)?temp1:temp2;
    
84. 
    
85. delete m1;
    
86. delete m2;
    
87. delete sm1;
    
88. delete sm2;
    
89. return;
    
90. }
    
91. 
    
92. }
    
93. int main()
    
94. {
    
95. int nums[] = {2, 3, 12, 108, 19, 34, 6, 7, 37, 28, 65, 26, 31, 68, 20, 5, 99};
    
96. 
    
97. int max, sm;
    
98. 
    
99. int size = sizeof(nums) / sizeof(int);
    
100. 
     
101. findMaxAndSubMax(nums, size, &max, &sm);
     
102. 
     
103. cout<<max<<endl;
     
104. cout<<sm<<endl;
     
105. 
     
106. 
     
107. return 0;
     
108. }

复制代码

CHLOE

楼上大哥的程序看的我头都晕啦!

（?羙‘妞

4楼和6楼都好复杂,只要排一下序不是就行了吗?

敬你永远遗忘

尽量利用语言已有的函数。
#include <iostream>
#include <algorithm>
using namespace std;
int main(int argc, char* argv[])
{
int intar[] = {5,8,2,6,4,3,11,10,9,7}, intSize = sizeof(intar)/sizeof(int);
sort(&intar[0], &intar[intSize]);
cout << intar[intSize-1] << " " << intar[intSize-2] << endl;
return 0;
}