有一17个元素的整形数组
寻找最大元素和次大元素

各位帮帮忙啊

先给它排序了,不就行了吗?

[此贴子已经被作者于2006-5-28 16:32:18编辑过]

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();
}

可以用list类，然后再sort();就可以了

[CODE]#include <iostream>
using namespace std;


void findMaxAndSubMax(int *pi, int size, int * m, int *sm)
{
if(size == 2)
{
if(pi[0]>pi[1])
{
*m = pi[0];
*sm = pi[1];
return;
}
*m = pi[1];
*sm = pi[0];
return;
}
else if(size == 3)
{
if(pi[0]>pi[1] && pi[1]>pi[2])
{
*m = pi[0];
*sm = pi[1];
return;
}
else if(pi[0]>pi[2] && pi[2]>pi[1])
{
*m = pi[0];
*sm = pi[2];
return;
}
else if(pi[1]>pi[0] && pi[0]>pi[2])
{
*m = pi[1];
*sm = pi[0];
return;
}
else if(pi[1]>pi[2] && pi[2]>pi[0])
{
*m = pi[1];
*sm = pi[2];
return;
}
else if(pi[2]>pi[0] && pi[0]>pi[1])
{
*m = pi[2];
*sm = pi[0];
return;
}
else if(pi[2]>pi[1] && pi[1]>pi[0])
{
*m = pi[2];
*sm = pi[1];
return;
}
}
else
{
int * m1 = new int;
int * sm1 = new int;
int * m2 = new int;
int * sm2 = new int;

int firstHalf = size/2;
int secondHalf = size - firstHalf;

findMaxAndSubMax(pi, firstHalf, m1, sm1);
findMaxAndSubMax(pi+firstHalf, secondHalf, m2, sm2);

int temp1, temp2;
if(*m1 > *m2)
{
*m = *m1;
temp1 = *m2;
}
else
{
*m = *m2;
temp1 = *m1;
}
temp2 = (*sm1>*sm2)?*sm1:*sm2;
*sm = (temp1>temp2)?temp1:temp2;

delete m1;
delete m2;
delete sm1;
delete sm2;
return;
}

}
int main()
{
int nums[] = {2, 3, 12, 108, 19, 34, 6, 7, 37, 28, 65, 26, 31, 68, 20, 5, 99};

int max, sm;

int size = sizeof(nums) / sizeof(int);

findMaxAndSubMax(nums, size, &max, &sm);

cout<<max<<endl;
cout<<sm<<endl;


return 0;
}[/CODE]

楼上大哥的程序看的我头都晕啦!

4楼和6楼都好复杂,只要排一下序不是就行了吗?

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;
}

觉得如果还不是很熟练算法思想的时候就多练习不用现有的排序函数，锻炼思维，但是如果掌握以后直接用algorithm或则stl可以简单解决问题，多好~！