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.

/*

The sorted array is rotated. Then there must be two sorted sub arrays. When do the binary search, we should discuss the target is in which part.

For example, given [4,5,6,7,0,1,2], there are two sorted parts, [4,5,6,7] and [0,1,2]

If nums[mid] < target, there will be three conditions:

1. nums[mid] and target are both in [0,1,2]

2. nums[mid] and target are both in [4,5,6,7]

3. nums[mid] is in [0,1,2] and target is in [4,5,6,7]

Then when do the binary search, the search scope can be reduced by making left = mid in condition 1,2 or right = mid in condition 3.

It is similar when nums[mid] > target.

time:O(logn), space:O(1)

*/

public class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int left = 0;
        int right = nums.length - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) {
                if (nums[mid] <= nums[right] && 
                    target > nums[right]) {
                    right = mid;
                } else {
                    left = mid;
                }
            } else if (nums[mid] > target) {
                if (nums[mid] > nums[right] && 
                    target <= nums[right]) {
                    left = mid;
                } else {
                    right = mid;
                }
            } else {
                return mid;
            }
        }
        if (nums[left] == target) {
            return left;
        }
        if (nums[right] == target) {
            return right;
        }
        return -1;
    }
}