算法思想:先相邻两个两个比较,较大的放入数组max[],较小的放入数组min[],然后从max[]数组求出最大,min[]数组求出最小即可。
比较n+[(n+1)/2] =1.5n次
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#include <iostream>
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#define n 11
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#define m ((n+1)/2)
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using namespace std;
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void main(void)
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![](http://www.cppblog.com/Images/OutliningIndicators/ContractedBlock.gif)
{
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int num[] =
{11,2,3,4,6,5,7,8,9,10,20};
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//int n = sizeof(num)/sizeof(num[0]);
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//int m = (n+1)/2;
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int max[m] , min[m];
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int k = 0, j = 0;
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if(n/2 != 0) max[m-1] = min[m-1] = num[n-1];
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for (int i=0; i < n-1; i = i+2)
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{
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if (num[i] >= num[i+1])
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{
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max[j++] = num[i];
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min[k++] = num[i+1];
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}
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else
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{
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max[j++] = num[i+1];
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min[k++] = num[i];
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}
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}
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for( i=0; i< m; i++)
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{
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cout << "max[" << i << "] = " << max[i] << "\t";
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cout << "min[" << i << "] = " << min[i] <<endl;
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}
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int MAX = max[0];
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int MIN = min[0];
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for ( j = 1; j < m; j++)
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{
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if (max[j] > MAX) MAX = max[j];
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if (min[j] < MIN) MIN = min[j];
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}
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cout << "MAX = " << MAX << ", MIN = " << MIN <<endl;
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}