1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 
 4 //例如12345 的旋轉數組34512   這類的左側第一個值大一等於右側   由於每個子數組是排好序的，所以用二分查找logn
 5 //例如01111 的旋轉數組11101  10111這種第一個值=中間值=最後值需要特殊處理   只能遍歷 n
 6 class Solution {
 7 public:
 8     int minNumberInRotateArray(vector<int> rotateArray) {
 9         int n = rotateArray.size();
10         if(n==0)
11             return 0;
 
12         int index1 = 0;
13         int index2 = n-1;
14         int indexMid = index1;
15         while(rotateArray[index1]>=rotateArray[index2])
16         {
17             if(index2-index1 == 1)//相差一個時，是查找截止條件
18             {
19                 indexMid = index2;
20                 break;
21             }
 
22             indexMid = (index1+index2)/2;
23             if(rotateArray[index1] == rotateArray[index2] && rotateArray[index1] == rotateArray[indexMid])//特殊處理
24             {
25                 return MinInOrder(rotateArray,index1,index2);
26             }
27             if(rotateArray[indexMid]>=rotateArray[index1])//中間值大於左側值，最小值在右側
28             {
29                 index1 = indexMid;
30             }
31             else if(rotateArray[indexMid] <=rotateArray[index2])//中間值小於右側值最小值在左側
32             {
33                 index2 = indexMid;
34             }
35         }
36         return rotateArray[indexMid];
37 
38     }
39     int MinInOrder(vector<int> rotateArray,int index1,int index2)
40     {
41         int result = rotateArray[index1];
42         for(int i=index1+1;i<=index2;i++)
43         {
44             if(rotateArray[i]<result)
45             {
46                 result = rotateArray[i];
47             }
48         }
49         return result;
50     }
51 };
52 int main()
53 {
54     vector<int> a;
55     a.push_back(3);
56     a.push_back(4);
57     a.push_back(5);
58     a.push_back(1);
59     a.push_back(2);
60     Solution s;
61     int result = s.minNumberInRotateArray(a);
62     cout<<"the min of 12345 is"<<result<<endl;
63 
64     vector<int> b;
65     b.push_back(1);
66     b.push_back(0);
67     b.push_back(1);
68     b.push_back(1);
69     b.push_back(1);
70     result = s.minNumberInRotateArray(b);
71     cout<<"the min of 01111 is"<<result<<endl;
72     return 0;
73 }

輸出結果截圖：

6、劍指offer--旋轉數組的最小數字