数组（4）：Search in Rotated Sorted Array II

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.

分析：
允许重复元素，则上一题中如果A[m]>=A[l]，那么[1,m]为递增序列的假设就不能成立了，比如[1,3,1,1,1]。
如果A[m]>=A[l]不能确定递增，那就把它拆分为两个条件：
（1）如果 A[m]>A[l]，则区间一定递增；
（2）如果 A[m]==A[l]确定不了，则i++，往下看一步即可。
代码

// 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;
   }
};