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

 1 #include <iostream>
 2 #include <algorithm>
 3 using namespace std;
 4
 5 double num[100];
 6 int n = 0;
 7
 8 // 用一个结构体来记录最大最小值
 9 struct Res
10 {
11     double maxs;
12     double mins;
13 };
14
15 struct Res MaxAndMin(int left,int right)
16 {
17     Res res,a,b;
18
19     if(left +1 >= right) //判断是否递归到底部
20     {
21         res.maxs = max(num[left],num[right]);
22         res.mins = min(num[left],num[right]);
23         return res;
24     }
25
26     int middle = (right + left) / 2;
27     a = MaxAndMin(left,middle);
28     b = MaxAndMin(middle+1,right);
29     res.maxs = max(a.maxs,b.maxs);
30     res.mins = min(a.mins,b.mins);
31     return res;
32 }
33 int main()
34 {
35     while(cin>>n)
36     {
37         //输入n个数
38         for(int i = 0; i < n; i++)
39         {
40             cin>>num[i];
41         }
42
43         //只有一个数时，直接输出
44         if(n == 1)
45         {
46             cout<<"最大值:"<<num[0]<<" 最小值:"<<num[0]<<endl;
47             continue;
48         }
49         //只有两个数时，比较这两个数，然后输出
50         if(n == 2)
51         {
52             if(num[0] > num[1])
53             {
54                 cout<<"最大值:"<<num[0]<<" 最小值:"<<num[1]<<endl;
55             }
56             else
57             {
58                 cout<<"最大值:"<<num[1]<<" 最小值:"<<num[0]<<endl;
59             }
60             continue;
61         }
62         //多个数时，分治的思想，将数组二分
63         int left = 0,right = n-1;
64         Res result;
65         result = MaxAndMin(left,right);
66         cout<<"最大值:"<<result.maxs<<" 最小值:"<<result.mins<<endl;
67     }
68
69     return 0;
70 }

 1 void FindMaxAndMinMethod4(int *pArr, int nStart, int nEnd, int &nMax, int &nMin)
 2 {
 3     if (nEnd - nStart <= 1)
 4     {
 5         if (pArr[nStart] > pArr[nEnd])
 6         {
 7             nMax = pArr[nStart];
 8             nMin = pArr[nEnd];
 9         }
10         else
11         {
12             nMax = pArr[nEnd];
13             nMin = pArr[nStart];
14         }
15         return;
16     }
17
18     int nLeftMax = 0;
19     int nLeftMin = 0;
20     int nRightMax = 0;
21     int nRightMin = 0;
22     FindMaxAndMinMethod4(pArr, nStart, nStart+(nEnd-nStart)/2, nLeftMax, nLeftMin);
23     FindMaxAndMinMethod4(pArr, nStart+(nEnd-nStart)/2+1, nEnd, nRightMax, nRightMin);
24
25     nMax = nLeftMax > nRightMax ? nLeftMax : nRightMax;
26     nMin = nLeftMin < nRightMin ? nLeftMin : nRightMin;
27 }