[LeetCode] Search in Rotated 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.

思路：本题思路和http://www.cnblogs.com/vincently/p/4122528.html类似。只是如果有重复的话就不能依靠与边界的比较确定递增的序列。例如，[1,3,1,1,1]，对于A[m]>=A[l]来说，[l,m]的假设就不成立。但是还是可以观察到，左边子数组的值都要大于等于右边子数组的值。将A[m]>=A[l]拆分为两种情况：A[m]>A[l],则区间[l,m]一定递增；如果 A[m]==A[l]确定不了，那就l++，往下看一步。

时间复杂度改变为O(n),空间复杂度O(1)

 class Solution {
 public:
     bool search(int A[], int n, int target) {
         int first = , last = n - ;
         while (first <= last) {
             const int mid = low + ((high - low) >> 1);
             if (A[mid] == target) return true;
             if (A[first] < A[mid]) {
                 if (A[first] <= target && target < A[mid])
                     last = mid - ;
                 else
                     first = mid + ;
             } else if (A[first] > A[mid]){
                 if (A[mid] < target && target <= A[last])
                     first = mid + ;
                 else
                     last = mid - ;
             } else {
                 first++;
             }
         }
         return false;
     }
 };