[leedcode 81] Search in Rotated Sorted Array II

Follow up for "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.

public class Solution {
    public boolean search(int[] nums, int target) {
        //注意利用本题的规律进行判断：
        //nums[0]是重点，它是前半段的最小值，后半段的最大值，因此mid需要与它判断大小，进而区分mid落在哪个范围
        //中值和start判断，如果大于start，说明start到mid是递增序列，只要判断target是否在这区间即可
        //如果小于start，说明mid到end是递增序列，只要判断target是否在这个区间即可
        //如果中值等于start（暗含nums[start]不等于target），因此此时start++；
        return findSearch(nums,0,nums.length-1,target);
    }
    public boolean findSearch(int[] nums,int start,int end,int target){
        if(start>end){
            return false;
        }
        int mid=(start+end)/2;
        if(nums[mid]==target) return true;
        if(nums[mid]>nums[start]){
            if(target<nums[mid]&&target>=nums[start]){
                end=mid-1;
            }else{
                start=mid+1;
            }
        }else if(nums[mid]<nums[start]){
            if(target>nums[mid]&&target<=nums[end]){
                start=mid+1;
            }else{
                end=mid-1;
            }
        }else start++;////////
         return findSearch(nums,start,end,target);
    }
}