33. Search in Rotated Sorted Array

Suppose a sorted array 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.

Related issue:81 Search in Rotated Sorted Array II

ATTENTION 1: since mid can equal equal to left, so >=.

ATTENTION2 : you need first focus on finding the portion of array that is ordered.

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

    }
}