算法4-1.1.29等值键

为 BinarySearch 类添加一个静态方法 rank (), 它接受一个键和一个整型有序数组(可能存在重复键)作为参数并返回数组中小于该键的元素数量,以及一个类似的方法 count () 来返回数组中等于该键的元素的数量。注意:如果1和j分别是rank(key,a)和count(key,a)的返回值,那么 a[i … i+j-1]就是数组中所有和key 相等的元素。

代码

public class Main {
    public static int rank(int[] a, int t, int lo, int hi, int depth){

        while (lo <= hi){
            int mid = lo + (hi-lo)/2;
            if (a[mid] < t){
                lo = mid+1;
            }else if (a[mid]>t){
                hi = mid-1;
            }else{
                return mid;
            }
            depth++;

        }
        return -1;
    }
    public static int count(int[] a, int key){
        int pos = rank(a,key);
        int localPos = pos;
        while (pos<a.length-1 && a[pos]==key){
            pos++;
        }
        return pos-localPos;
    }
    public static int rank(int[] a, int key){
        int lo = 0, hi = a.length, mid;
        while (lo < hi){
            mid = lo+(hi-lo)/2;
            if (a[mid] < key) mid = rank(a, key, mid+1, hi, 0);
            else if (a[mid] > key) mid =  rank(a, key, lo, mid-1,0);

            while (mid >= 0 && a[mid] == key){
                mid --;
            }
            return mid+1;
//                for (int i = mid; i >= 0; i --){
//                    if (a[i] < key) return i+1;
//                }
            }

        return -1;

    }
    public static void main(String[] args) {
        int[] a = { 1, 2, 3, 4, 4, 4, 4, 7, 7, 8,8, 9, 12 };
        Arrays.sort(a);
        System.out.println(rank(a, 7));
        System.out.println(count(a, 7));
    }
}

答案

7
2

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