[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;
    }
}