[topcoder]BinarySearchable

http://community.topcoder.com/stat?c=problem_statement&pm=5869&rd=8078

这道题有点意思，思考理解后，就是找数组中的数A[i]满足：左边的都比它小，右边的都比它大。所以从左向右扫，不断记录更新max值，对不满足的标记false；然后从右向左扫就行了。

public class BinarySearchable

{

    public int howMany(int[] sequence)

    {

        if (sequence.length == 0) return 0;

        int len = sequence.length;

        int max = sequence[0];

        boolean valid[] = new boolean[len];

        for (int i = 0; i < len; i++) valid[i] = true;

        for (int i = 1; i < len; i++)

        {

            if (sequence[i] > max)

                max = sequence[i];

            else // < previous one, in-valid

                valid[i] = false;

        }

        int min = sequence[len-1];

        for (int i = len-2; i>=0; i--)

        {

            if (sequence[i] < min)

                min = sequence[i];

            else

                valid[i] = false;

        }

        int cnt = 0;

        for (int i = 0; i < len; i++)

        {

            if (valid[i]) cnt++;

        }

        return cnt;

    }

}