leetcode 之Search in Rotated Sorted Array（四）

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.

如果循环数组里有重复元素，则根据A[m]>=A[l]是无法判断出（m,l）之间是升序的。因此，需将大于和等于分开来考虑。

只需在等于时，将重复元素跳过再比较即可。

 int searchRotateSA(int A[], int n,int target)
 {
     int first = , last = n;
     while (first!=last)
     {
         int mid = first + (last - first) / ;
         if (A[mid] = target)
         {
             return mid;
         }
         else if (A[first]<A[mid])//判断大小顺序
         {
             if (A[first] <= target&&target <= A[mid])
                 last = mid;
             else
                 first = mid + ;
         }
         else if (A[first]>A[mid])
         {
             if (A[first] >= target&& A[mid] <= target)
                 last = mid;
             else
                 first = mid + ;

         }
         else first++;
     }

     return -;
 }