题目描述：
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.
思路：
这个题目与Search in Rotated Sorted Array基本是一样的，在旋转过的排序好的数组中进行二分查找，区别只是在于这个题目的输入可能包含重复元素。
本题解法基本与Search in Rotated Sorted Array I相同，区别只是需要跳过重复元素。
这里是Search in Rotated Sorted Array的解法：
<link>
而对于本题，需要在得到中间索引m后跳过左右重复元素，并重新计算m：
while(l < m && nums[l] == nums[m]){
l++;
}
while(r > m && nums[r] == nums[m]){
r--;
}
m = (l + r) / 2;
实现代码：public class Solution {
public bool Search(int[] nums, int target) {
if(nums.Length == 0)
{
return false;
}
if(nums.Length == 1)
{
return nums[0] == target ;
}
var l = 0;
var r = nums.Length - 1;
while(l < r - 1){
var m = (l + r) / 2;
while(l < m && nums[l] == nums[m]){
l++;
}
while(r > m && nums[r] == nums[m]){
r--;
}
m = (l + r) / 2;
if(nums[m] == target || nums[r] == target || nums[l] == target){
return true;
}
if(nums[m] > nums[l] && nums[m] > nums[r]){
if(target < nums[r]){
l = m;
}
else {
if(target > nums[m]){
l = m;
}
else{
r = m;
}
}
}
else if(nums[m] < nums[l] && nums[m] < nums[r]){
if(target > nums[l]){
r = m;
}
else{
if(target > nums[m]){
l = m;
}
else{
r = m;
}
}
}
else if(nums[m] > nums[l] && nums[m] < nums[r]){
if(target > nums[m]){
l = m;
}
else {
r = m;
}
}
}
if(target == nums[l] || target == nums[r]){
return true;
}
return false;
}
}