33. Search in Rotated Sorted Array

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

// array 可以是有序的；若是无序，则二分后，其中一半必然有序，从有序的部分进行判断二分走向

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