最小最大数

#include<iostream>
#include<vector>
#include<ctime>
#define N 1000
using namespace std;

int count1,count2,count3;

//算法1,2来自编程珠玑
void minmax(vector<int> a){
    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];
        }
        if(a[i]>max){
            max=a[i];    
        }
        count1 += 2;
    }
    cout<<"min,max="<<min<<","<<max<<endl;
}

void minmax2(vector<int> a){
    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];
                        ++count2;        
        }
        else{
                        if(a[i]>max){
                    max=a[i];
                        }
                        count2 += 2;
        }
    }
    cout<<"min,max="<<min<<","<<max<<endl;
}

void minmax3(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];
            }
        }
        count3 += 3;
    }
    cout<<"min,max="<<min<<","<<max<<endl;
}

int main(){
    vector<int> a;
    for(int i=0;i<N;i++){
        a.push_back(i);
    }

    for(int i=0;i<N;i++){
        swap(a[i],a[rand()%(N-i)+i]);
    }

    //clock_t  st,fi,st2,fi2,st3,fi3;

    //st = clock();
    minmax(a);
    //fi = clock();
    cout<<"minmax比较次数 :"<<count1<<endl;
        cout<<endl;

    //st2 = clock();
    minmax2(a);
    //fi2 = clock();
    cout<<"minmax2比较次数"<<count2<<endl;
        cout<<endl;

    //st3 = clock();
    minmax3(a);
    //fi3 = clock();
    cout<<"minmax3比较次数"<<count3<<endl;
    return 0;
}                

算法1,2来自<<编程珠玑续>>第一章第一题,2是对1的优化,因为如果a[i]<min,那么a[i]不可能大于max.但这样做优化幅度并不大,假设a中元素均匀散布,Knuth证明a[i]<min成立的次数为H_N-1,其中H_N=1+1/2+1/3+..+1/N,即为第N个调和数.对于N=1000,其期望值为6.485.

算法3来自<<算法导论>>,思路如下:

1. Pick 2 elements(a, b), compare them. (say a > b)
2. Update min by comparing (min, b)
3. Update max by comparing (max, a)

比较次数为3N/2证明如下:

This way you would do 3 comparisons for 2 elements, amounting to 3N/2 total comparisons for N elements.

程序运行结果:(N=1000)

虽然比较次数有区别,但当N较大(N=1000000)时,三个算法的实际运行时间并无太大差别.当N更大(N=100000000)时,可以明显看出算法3最慢,算法2最快,算法1次之.

N=1000000　　　　　　　　　　　　N=100000000