二分查找及相关问题

Posted 2020-08-05 BBJ

1.旋转数组中找最小值

顺序查找时间复杂度为O(n),二分查找时间复杂度为O(logn)

 1 // rotateArrMin.cpp : 定义控制台应用程序的入口点。
 2 //
 3 
 4 #include "stdafx.h"
 5 #include <iostream>
 6 using namespace std;
 7 
 8 void MinInOrder(int *arr,int begin,int end)
 9 {
10     int result = arr[begin];
11     for(int i = begin+1; i < end;i++)
12         result = min(result,arr[i]);
13     cout<<result<<endl;
14 }
15 
16 void Min(int* arr,int len)
17 {
18     if(arr == NULL || len <= 0)
19         return;
20 
21     int begin = 0;
22     int end = len - 1;
23     int middle = begin;
24     int result = 0;
25     //强middle初始化为begin,在原数组中也能返回第一个数即最小值,(此时旋转了0)
26     while(begin <= end)
27     {
28         if(end - begin == 1)
29         {
30             result = arr[end];
31             break;
32         }
33         middle = (begin + end)>>1;
34 
35         if(arr[begin] == arr[middle] && arr[middle] == arr[end])
36             MinInOrder(arr,begin,end);
37         //处理01111的旋转，11101等
38 
39         if(arr[middle] >= arr[begin])
40             begin = middle;
41         else if(arr[middle] <= arr[end])
42             end = middle;
43     }
44     cout<<result<<endl;
45 }
46 
47 int _tmain(int argc, _TCHAR* argv[])
48 {
49     int arr[] = {3,4,5,1,2};
50     int len = 5;
51     Min(arr,len);
52     system("pause");
53     return 0;
54 }

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