81. Search in Rotated Sorted Array II

Follow up for 33 Search in Rotated Sorted Array:

What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

ATTENTION: when nums[mid] == nums[left], you cannot gurantee the shape of array.

public class Solution {
    public boolean 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 true;
            if(nums[mid] > nums[left]){ 
                if(nums[left] <= target && target < nums[mid]){
                    right = mid-1;
                }else{
                    left = mid +1;
                }
            }else if (nums[mid] < nums[left]){
                if(nums[mid] <= target && target <= nums[right]){
                    left = mid +1;
                }else{
                    right = mid -1;
                }
            }else{
                left++;// move a step ahead. cause  there is no gaurantee the target will show on which part. cause the shap of first part[left, mid] is not decided.
            }
        }
        return false;
    }
}