旋转数组的最小数字

 1 /**
 2  * 旋转数组的最小数字
 3  * 
 4  * @author Administrator
 5  * 
 6  *         测试用例： null 1 1,2,3,4,5 3,4,5,1,2 1,0,1,1,1 1,1,1,0,1 2,3,4,5,1,2,2,2
 7  * 
 8  *         时间复杂度：O(logn)
 9  */
10 public class Main {
11 
12     public static void main(String args[]) {
13 
14         int[] array = { 2, 3, 4, 5, 1, 2, 2, 2 };
15 
16         Main main = new Main();
17 
18         System.out.println("min num: " + main.getMinNum(array));
19     }
20 
21     private Integer getMinNum(int[] array) {
22 
23         if (null == array || 0 == array.length) {
24             return null;
25         }
26 
27         if (array.length == 1) {
28             return array[0];
29         }
30 
31         int head = 0;
32         int tail = array.length - 1;
33         int middle = 0;
34 
35         if (array[head] < array[tail]) {
36             return array[head];
37         }
38 
39         while (tail - head > 1) {
40 
41             middle = (head + tail) / 2;
42 
43             // 头 尾 中间元素值相同，无法判断指针的移动方向，转为顺序查找
44             if (array[head] == array[middle] && array[middle] == array[tail]) {
45                 return minOrder(array, head, tail);
46             }
47 
48             // 中间元素值大于头元素值，头指针移动到中间位置
49             if (array[head] <= array[middle]) {
50                 head = middle;
51                 continue;
52             }
53 
54             // 中间元素值小于尾元素值，尾元素移动到中间位置
55             if (array[tail] >= array[middle]) {
56                 tail = middle;
57                 continue;
58             }
59         }
60 
61         return array[tail];
62     }
63 
64     private Integer minOrder(int[] array, int head, int tail) {
65 
66         int minNum = array[head];
67 
68         for (int i = head; i <= tail; i++) {
69             if (array[i] < minNum) {
70                 minNum = array[i];
71             }
72         }
73 
74         return minNum;
75     }
76 }