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.

Solution 1 Binary search

public static boolean search(int[] nums, int target) {
int start = 0, end = nums.length - 1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target)
return true; // return mid in Search in Rotated Array I
if (nums[mid] > nums[start]) { // left half is sorted
if (target >= nums [start] && target < nums[mid])
end = mid - 1;
else
start = mid + 1;
} else if (nums[mid] < nums[start]) { // right half is sorted
if (target > nums[mid] && target <= nums[end])
start = mid + 1;
else
end = mid - 1;
} else
start++;
}
return false;
}