解题思路

n个大小的整数数组 思路1：遍历整个数组分别得出最大值和最小值，比较次数为2n 思路2：取两个数进行比较，大的与当前最大值比较，小的与当前最小值比较，则没两个数要比较3次，比较次数为1.5n

c++代码

//取两个数进行比较，小的数跟当前最小值比较，大的数跟当前最大值比较 void GetMinMax(int* buffer,int& bufferLen,int& max, int& min) { if(buffer == NULL || bufferLen <= 0) { cout<<"buffer is empty or wrong"<<endl; return; } int i = 0; int j = bufferLen -1; min = buffer[0]; max = buffer[0]; while(i <= j) { if(buffer[i] < buffer[j]) { if(buffer[i] < min) { min = buffer[i]; } if(buffer[j] > max) { max = buffer[j]; } } else { if(buffer[j] < min) { min = buffer[j]; } if(buffer[i] > max) { max = buffer[i]; } } i++; j--; } }

测试代码

int _tmain(int argc, _TCHAR* argv[]) { int buffer[10] = {9,6,1,7,10,2,4,5,8,-1}; int bufferLength = 10; int min = 0; int max = 0; GetMinMax(buffer,bufferLength,max,min); cout<<"min = "<<min<<" max = "<<max<<endl; return 0; }