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 = 0, last = n;
    while (first!=last)
    {
        int mid = first + (last - first) / 2;
        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 + 1;
        }
        else if (A[first]>A[mid])
        {
            if (A[first] >= target&& A[mid] <= target)
                last = mid;
            else
                first = mid + 1;
            
        }
        else first++;
    }

    return -1;
}