https://leetcode.com/problems/search-in-rotated-sorted-array/description/
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
time: o(logn) space: o(1)
与Find Minimum in Rotated Sorted Array类似。因为rotate, 所以不能直接用Binary Search, 需要进行 二次判定。
case 1: nums[mid] == target, return mid.
case 2: nums[mid] < nums[r]. 此时说明 右侧是正常升序，没有rotation, 若是这种情况下 target > nums[mid] && target <= nums[r] 就在右侧查找，其他情况左侧查找。
case 3: nums[mid] > nums[r]. 此时说明 左侧是正常升序，没有rotation, 若是这种情况下 target < nums[mid] && target >= nums[l] 就在左侧查找，其他情况右侧查找
note here since there is no duplicate and the left + 1 = mid will exit, there wouldnt be a case that the nums[left]= nums[mid] or nums[mid] == num[right]
make sure also checks the LC.81
1 public int search(int[] nums, int target) {
2 if (nums == null || nums.length ==0 ) return -1 ;
3 int left = 0, right = nums.length -1 ;
4 while(left + 1 < right){
5 int mid = left + (right-left)/2;
6 if (nums[mid] == target) return mid ;
7 //break into two parts: note, there is no duplicate
8 //first half
9 if (nums[left]< nums[mid] ){
10 if (target<=nums[mid] && nums[left] <=target){
11 right = mid ;
12 }else {
13 left = mid ;
14 }
15 }
16 //second half:注意判断顺序进行改变
17 else if (nums[mid] < nums[right]){
18 if (nums[mid]<=target && target <= nums[right]){
19 left = mid ;
20 }else {
21 right = mid;
22 }
23 }
24 }
25 //post processing for the left and right
26 if (nums[left] == target){
27 return left ;
28 }
29 if(nums[right] == target){
30 return right;
31 }
32 return -1 ;
33 }
