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 }
View Code
 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 }
View Code

 

posted @ 2016-01-31 17:19  Gladitor  阅读(520)  评论(0)  编辑  收藏  举报