Question：
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
Write a function to determine if a given target is in the array.
The array may contain duplicates.
Tips：
给定一个数组，该数组是由一个有序的数组经过旋转（即将前面一段数字接到整个数组之后）得到的。判断target是否存在于该数组之中。
本题为33题升级版本，数组中的数字可以出现重复，如果target存在，返回他的true不存在则返回false。
思路：
本题与33提相似，但是由于数组中存在重复数字，可能会出现low high mid三个数字均相等的情况，这时为了跳出相等数字，需要low++或者high--;
代码：
public boolean search(int[] nums, int target) {
if (nums == null)
return false;
int low = 0;
int len = nums.length;
int high = len - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == target)
return true;
if (nums[low] < nums[mid] || nums[mid]>nums[high]) {
if (target < nums[mid] && target >= nums[low]) {
high = mid - 1;
} else
low = mid + 1;
}else if(nums[mid]<nums[high] || nums[low]>nums[mid]){
if(target<=nums[high] && target>nums[mid]){
low=mid+1;
}else{
high=mid-1;
}
}
else{
low++;
}
}
return false;
}