# 33. Search in Rotated Sorted Array (Medium)

https://leetcode.com/problems/search-in-rotated-sorted-array/

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```

## Solutions

``````class Solution {
public int search(int[] nums, int target) {
int len = nums.length;

if (nums == null || len == 0) {
return -1;
}

if (len == 1) {
return nums == target ? 0 : -1;
}

int left = 0;
int right = len - 1;

// first find the rotated index
int rindex = 0;

while (left < right) {
int mid = (left + right) / 2;

if (nums[mid] > nums[mid + 1]) {
rindex = mid + 1;
break;
}

if (nums[mid] > nums[right]) {
left = mid;
}

if (nums[mid] < nums[right]) {
right = mid;
}
}

if (rindex == 0) { // not rotated
left = 0;
right = len - 1;
} else if (nums[len - 1] < target) { // on left part
left = 0;
right = rindex - 1;
} else { // on right part, rindex != 0 && nums[len-1] > target
left = rindex;
right = len - 1;
}

while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
return mid;
}

if (nums[mid] > target) {
right = mid - 1;
} else {
left = mid + 1;
}
}

return -1;
}
}
``````