最小最大数

第二个算法的时间复杂度为O(n),比较次数为3N/2

#include<iostream>
#include<vector>
#include<ctime>

using namespace std;

void minmax(vector<int> a){//来自编程珠玑
    int min=a[0];
    int max=a[0];
    int n=a.size();
    for(int i=1;i<n;i++){
        if(a[i]<min){
            min = a[i];
        }
        else if(a[i]>max){
            max=a[i];
        }
    }
    cout<<"min,max="<<min<<","<<max<<endl;
}

void minmax2(vector<int> a){//来自算法导论
    int n = a.size();
    int min=a[0];
    int max;
    int k;
    if(n%2!=0){
        max = a[0];
        k=1;
    }
    else{
        max = a[1];
        k=2;
    }
    for(;k<n;k += 2){
        if(a[k]<a[k+1]){
            if(a[k]<min){
                min = a[k];
            }
            if(a[k+1]>max){
                max = a[k+1];
            }
        }
        else{
            if(a[k+1]<min){
                min = a[k+1];
            }
            if(a[k]>max){
                max = a[k];
            }
        }
    }
    cout<<"min,max="<<min<<","<<max<<endl;

}

int main(){
    vector<int> a{12,42,23,3,45,99,21,76};
    clock_t  st,fi,st2,fi2;
    st = clock();
    minmax(a);
    fi = clock();
    cout<<"minmax:"<<(fi-st)<<endl;
    st2 = clock();
    minmax2(a);
    fi2 = clock();
    cout<<"minmax:"<<(fi2-st2)<<endl;
    return 0;
}