查找数组最大值最小值:时间复杂度最多为：o(3n/2)

#include <iostream>using namespace std;/*查找数组最大值最小值:时间复杂度最多为：o(3n/2)*//*基本思想：当n=2k+1，将让max和min都等于第一个元素，让后将剩余的元素两两比较，将大值再与max比较，如果最大值大于max，则让max等与该最大值，将两个数中的小值再与min比较，若小于min，则让min等于该小值。假定n=2k,设置最大值和最小值为第一、二个元素，即Max=A[0],Min=A=[1];将n个数据按2个一组进行分组，n/2组数各自进行比较得到较大值P和较小值q；将Max、Min分别和每组中的较大值、较小值比较，若Max>p,则Max=p;同理，若Min<q,则Min=q;如此循环n/2次即可。 每组比较三次，一共比较n/2次，复杂度为o(3n/2)*/void MaxMin(int arr[], int n, int &max, int &min){int i;/*元素个数为基数时*/if (n%2 == 1){max = min = arr[0];for (i=1; i<n; i+=2){if (arr[i] > arr[i+1]){max = (arr[i] > max) ? arr[i] : max;min = (arr[i+1] < min) ? arr[i+1] : min;}else{max = (arr[i+1] > max) ? arr[i+1] : max;min = (arr[i] < min) ? arr[i] : min;}}}else /*偶数情况*/{max = (arr[0] > arr[1]) ? arr[0] : arr[1];min = (arr[0] < arr[1]) ? arr[1] : arr[0];for (i=2; i<n; i+=2){if (arr[i] > arr[i+1]){max = (arr[i] > max) ? arr[i] : max;min = (arr[i+1] < min) ? arr[i+1] : min;}else{max = (arr[i+1] > max) ? arr[i+1] : max;min = (arr[i] < min) ? arr[i] : min;}}}}int main(){int arr[] = {2, 3, 54, 6, 0, 5, 1, -3};int size = sizeof(arr) / sizeof(int);int max, min;MaxMin(arr, size, max, min);cout<<max<<" "<<min<<endl;} 

 

 

 

#include <iostream>using namespace std;//函数功能：递归求数组的最大元素和最小元素//参数：arr为数组，low为下标，high最后一个元素的后一个位置， void MaxMin(int arr[], int low, int high, int& max, int &min);int main(){int arr[] = {1, 2, 3, 4, 5};int n = sizeof(arr) / sizeof(int);int max, min;MaxMin(arr, 0, n, max, min);cout<<"max = "<<max<<endl;cout<<"min = "<<min<<endl;return 0;}void MaxMin(int arr[], int low, int high, int& max, int &min){if(low == high - 1) //一个元素 max = min = arr[low];else if(low == high -2)//两个元素 {if(arr[low] > arr[low + 1]){max = arr[low];min = arr[low + 1];}else{max = arr[low + 1];min = arr[low];}}else //进行递归 {int mid = (low + high) / 2;int max1 = 0;int min1 = 0;MaxMin(arr, low, mid, max, min);MaxMin(arr, mid, high, max1, min1);if(max1 > max)max = max1;if(min1 < min)min = min1;}}