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算法练习(22):Java基础:引用,别名,二分法查找(1.2.8-1.2.9)

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算法之路
2017.09.17 13:32* 字数 460

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算法(第4版)

知识点

  • 引用
  • 别名
  • 非主流 二分法查找的实现

题目

1.2.8设 a[] 和 b[] 均为长数百万的整型数组。以下代码的作用是什么?有效吗?
int[] t = a; a = b; b = t;


1.2.8 Suppose that a[] and b[] are each integer arrays consisting of millions of integers. What does the follow code do? Is it reasonably efficient?
int[] t = a; a = b; b = t;
Answer. It swaps them. It could hardly be more efficient because it does so by copying references, so that it is not necessary to copy millions of elements.

答案

作用就是交换两个数组。
在JAVA 中,数组变量实际是数组的一个引用(类似于指针),交换两个引用的效率与数组大小无关,都是常数时间的。

题目

1.2.9 修改 BinarySearch(请见 1.1.10.1 节中的二分查找代码),使用 Counter 统计在有查找中被检查的键的总数并在查找全部结束后打印该值。
提示:在 main() 中创建一个 Counter 对象并将它作为参数传递给 rank()


1.2.9 Instrument BinarySearch(page47) to use a Counter to count the total number of keys examined during all searches and then print the total after all searches are com- plete. Hint : Create a Counter in main() and pass it as an argument to rank().

答案

Counter.java

public class Counter {

    
    public int counter;
    
}

BinarySearchCounter.java

public class BinarySearchCounter {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        int[] numArray = { 1, 2, 3, 4, 67, 88, 89, 101, 222, 788, 999 };
        Counter counter = new Counter();
        int index = rank(222, numArray, counter);

        System.out.println("index: " + index + "\ncouter:" + counter.counter);
    }

    public static int rank(int t, int[] array, Counter counter) {

        int lo = 0;
        int hi = array.length - 1;
        int mid = (lo + hi) / 2;

        while (t != array[mid]) {
            counter.counter++;
            if (t < array[mid]) {

                if (hi == mid) {
                    return -1;
                }
                hi = mid;
            } else if (t > array[mid]) {
                if (lo == mid) {
                    return -1;
                }
                lo = mid;
            } else {
                return mid;
            }

            mid = (lo + hi) / 2;
        }

        return mid;
    }

}

代码索引

BinarySearchCounter.java

视频讲解

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