/
BinarySearchRecursion.java
52 lines (45 loc) · 1.73 KB
/
BinarySearchRecursion.java
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package com.kyson.chapter1.section1;
/**
* 1.1.22 使用1.1.6.4 中的 rank()递归方法重新实现 BinarySearch 并跟踪该方法的调用。每当该
* 方法被调用时，打印出它的参数 lo 和 hi 并按照递归的深度缩进。提示 :为递归方法加一个参数来保存
* 递归的深度。
* <p>
* 1.1.22 Write a version of Binary Search that uses the recursive rank() given
* on page 25 and traces the method calls. Each time the recursive method is
* called, print the argument values lo and hi, indented by the depth of the
* recursion. Hint: Add an argument to the recursive method that keeps track
* of the depth.
*/
public class BinarySearchRecursion {
public static int rank(int key, int[] a, int lo, int hi) {
if (lo > hi)
return -1;
int mid = lo + (hi - lo) / 2;
if (key < a[mid])
return rank(key, a, lo, mid - 1);
else if (key > a[mid])
return rank(key, a, mid + 1, hi);
else
return mid;
}
public static int rank(int key, int[] a) {
return rank(key, a, 0, 16, 1);
}
public static int rank(int key, int[] a, int lo, int hi, int deep) {
if (hi < lo) return -1;
int mid = lo + (hi - lo) / 2;
for (int i = 0; i < deep; i++)
System.out.print(" ");
System.out.println("lo: " + lo + " hi: " + hi);
if (key < a[mid])
return rank(key, a, lo, mid - 1, deep + 1);
else if (key > a[mid])
return rank(key, a, mid + 1, hi, deep + 1);
else
return mid;
}
public static void main(String[] args) {
int[] array = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16};
rank(10, array);
}
}