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.

与Search in Rotated Sorted Array的区别在于，当nums[start]
== nums[middle]时，需要将start往后移动。

class Solution {
public:
bool search(vector<int>& nums, int target) {
int start = 0;
int end = nums.size();
while(start < end)
{
int middle = (start+end) / 2;
if(target == nums[middle])
return true;
if(nums[start] < nums[middle])
{
if(target >= nums[start] && target <= nums[middle])
{
end = middle;
}
else
{
start = middle+1;
}
}
else if(nums[start] > nums[middle])
{
if(target >= nums[middle] && target <= nums[end-1])
{
start = middle + 1;
}
else
{
end = middle;
}

}
else
start++;
}
return false;

}
};