Difficulty: Medium
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.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1```
#### Solution
Language: **Java**
```java
class Solution {
public int search(int[] nums, int target) {
int low = 0;
int high = nums.length - 1;
while (low <= high) {
int mid = (low + high) / 2;
if (nums[mid] == target) {
return mid;
}
if (nums[mid] >= nums[low]) { //说明low到mid这一段是递增的，不会经过对称线
if (target >= nums[low] && target <= nums[mid]) { // 如果 target 在 这之间，继续二分即可
int result = Arrays.binarySearch(nums, low, mid, target);
return result >= 0 ? result : -1;
} else {
low = mid + 1;
}
} else {
if (target >= nums[mid] && target <= nums[high]) { // 如果 target 在 这之间，继续二分即可
int result = Arrays.binarySearch(nums, mid + 1, high + 1, target);
return result >= 0 ? result : -1;
} else {
high = mid - 1;
}
}
}
return -1;
}
}
