题目：

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.

题意：

继续上一题，如果有重复元素出现，那么该如何处理？还是如同上一题那么处理，采用二分搜索，然后考虑出现了重复的，那么就采用顺序遍历，知道找到了，否则就返回false.

public static boolean search(int[] nums, int target) {if(nums.length == 0 || nums == null)return false;int low = 0, high = nums.length - 1, mid;while(low <= high) {mid = low + (high - low) / 2;if(nums[mid] == target)return true;else if(nums[mid] > nums[low]) {if(target >= nums[low] && target < nums[mid])high = mid - 1;elselow = mid + 1;} else if(nums[mid] < nums[low]){if(target > nums[mid] && target <= nums[high])low = mid + 1;elsehigh = mid - 1;}else       //如果是相等的情况，也就是说又重复，那么考虑顺序遍历{low++;}   }return false;}