1 public int findMin(int[] nums)
2 {
3 return findMin(nums, 0, nums.length - 1);
4 }
5 public int findMin(int[] nums, int left, int right)
6 {
7 int mid = (left + right)/2;
8 if(left == right)
9 {
10 return nums[left];
11 }
12 if(nums[left] < nums[right])
13 {
14 return nums[left];
15 }
16 if((left - right) == 1)//这一步必须要有，因为会涉及到left+1，right-1，例如序列为11.
17 {
18 return Math.min(nums[left], nums[right]);
19 }
20 if(nums[left] == nums[mid] && nums[mid] == nums[right])
21 {
22 return findMin(nums, left + 1, right - 1);
23 }
24 else if(nums[left] <= nums[mid])
25 {
26 return findMin(nums, mid+1, right);
27 }
28 else
29 {
30 return findMin(nums, left, mid);
31 }
32 }
33 //迭代方法
34 public int findMin2(int[] nums, int left, int right)
35 {
36 while(left <= right)
37 {
38 int mid = (left + right)/2;
39 if(left == right)
40 {
41 return nums[left];
42 }
43 if(nums[left] < nums[right])
44 {
45 return nums[left];
46 }
47 if(nums[right] == nums[mid] && nums[mid] == nums[right])
48 {
49 left ++;
50 right --;
51 }
52 else if(nums[left] <= nums[mid])
53 {
54 left = mid + 1;
55 }
56 else
57 {
58 right = mid;
59 }
60 }
61 return nums[left];
62 }
