数组（4）：Search in Rotated Sorted Array II

describe:

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.

Analysis:
Repeated elements are allowed, then if A[m]>=A[l] in the previous question, then the assumption that [1,m] is an increasing sequence cannot be established, such as [1,3,1,1,1 ].
If A[m]>=A[l] cannot be determined to increase, then split it into two conditions:
(1) If A[m]>A[l], the interval must increase;
(2) If A [m]==A[l] can not be determined, then i++, just look down one step.
code

// LeetCode, Search in Rotated Sorted Array II
// 时间复杂度 O(n)，空间复杂度 O(1)
class Solution {
public:
bool search(const vector<int>& nums, int target) {
    int first = 0, last = nums.size();
    while (first != last) {
          const int mid = first + (last - first) / 2;
          if (nums[mid] == target)
          return true;
         if (nums[first] < nums[mid]) {
            if (nums[first] <= target && target < nums[mid])
            last = mid;
            else
            first = mid + 1;
         } else if (nums[first] > nums[mid]) {
            if (nums[mid] < target && target <= nums[last-1])
            first = mid + 1;
            else
            last = mid;
         } else
           //skip duplicate one
          first++;
     }
     return false;
   }
};