[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);
}
}