Search in Rotated Sorted Array II

跟1 差不多，就是多了个一旦相等，就挪边缘的条件， 参考http://blog.csdn.net/linhuanmars/article/details/20588511

public class Solution {

    public boolean search(int[] A, int target) {

        

        if(A==null || A.length==0)

            return false;

       

        int l = 0;

        int r = A.length-1;

        while(l<=r)

        {

            int m = (l+r)/2;

            if(target == A[m])

                return true;

            if(A[m]<A[r])

            {

                if(target>A[m] && target<=A[r])

                    l = m+1;

                else

                    r = m-1;

            }

            else if(A[m]>A[r])

            {

                if(target>=A[l] && target<A[m])

                    r = m-1;

                else

                    l = m+1;                    

            }

            else r--;

        }

        return false;

    }

}