分治法解决寻找数组中最大最小值的问题

输入: 数组A[i,…,j]

输出:数组A[i,…,j]中的max和min  

1. If  j-i+1 =1   Then 输出A[i],A[i],算法结束 

2. If  j-i+1 =2   Then  

3.      If  A[i]< A[j]  Then输出A[i],A[j];算法结束  

4. k<--(j-i+1)/2  

5. m1,M1<--MaxMin(A[i:k]);  

6. m2,M2 <--MaxMin(A[k+1:j]);  

7. m <--min(m1,m2);  

8. M <--min(M1,M2);  

9. 输出m,M

复杂度为3/2n-1

#include <iostream>
#include <algorithm>
using namespace std;

double num[100];
int n = 0;

// 用一个结构体来记录最大最小值
struct Res
{
    double maxs;
    double mins;
};

struct Res MaxAndMin(int left,int right)
{
    Res res,a,b;

    if(left +1 >= right) //判断是否递归到底部
    {
        res.maxs = max(num[left],num[right]);
        res.mins = min(num[left],num[right]);
        return res;
    }

    int middle = (right + left) / 2;
    a = MaxAndMin(left,middle);
    b = MaxAndMin(middle+1,right);
    res.maxs = max(a.maxs,b.maxs);
    res.mins = min(a.mins,b.mins);
    return res;
}
int main()
{
    while(cin>>n)
    {
        //输入n个数
        for(int i = 0; i < n; i++)
        {
            cin>>num[i];
        }

        //只有一个数时，直接输出
        if(n == 1)
        {
            cout<<"最大值:"<<num[0]<<" 最小值:"<<num[0]<<endl;
            continue;
        }
        //只有两个数时，比较这两个数，然后输出
        if(n == 2)
        {
            if(num[0] > num[1])
            {
                cout<<"最大值:"<<num[0]<<" 最小值:"<<num[1]<<endl;
            }
            else
            {
                cout<<"最大值:"<<num[1]<<" 最小值:"<<num[0]<<endl;
            }
            continue;
        }
        //多个数时，分治的思想，将数组二分
        int left = 0,right = n-1;
        Res result;
        result = MaxAndMin(left,right);
        cout<<"最大值:"<<result.maxs<<" 最小值:"<<result.mins<<endl;
    }

    return 0;
}

void FindMaxAndMinMethod4(int *pArr, int nStart, int nEnd, int &nMax, int &nMin)  
{  
    if (nEnd - nStart <= 1)  
    {  
        if (pArr[nStart] > pArr[nEnd])  
        {  
            nMax = pArr[nStart];  
            nMin = pArr[nEnd];  
        }  
        else  
        {  
            nMax = pArr[nEnd];  
            nMin = pArr[nStart];  
        }  
        return;  
    }  
  
    int nLeftMax = 0;  
    int nLeftMin = 0;  
    int nRightMax = 0;  
    int nRightMin = 0;  
    FindMaxAndMinMethod4(pArr, nStart, nStart+(nEnd-nStart)/2, nLeftMax, nLeftMin);  
    FindMaxAndMinMethod4(pArr, nStart+(nEnd-nStart)/2+1, nEnd, nRightMax, nRightMin);  
      
    nMax = nLeftMax > nRightMax ? nLeftMax : nRightMax;  
    nMin = nLeftMin < nRightMin ? nLeftMin : nRightMin;    
}