此代码生成一个随机数,按升序对其进行排序,并进行二分搜索以查找目标值。我的问题是如何修改此代码以找到给定目标的最大索引。例如数组有 { 1, 2 , 3, 5, 5, 5, 5},目标是 5,所以输出应该是 6 而不是 3。谢谢。
import java.util.*;
public class Sort
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
System.out.print("How many numbers do you want? ");
int howMany = in.nextInt();
int [] myArray = getSortedRandomArray(howMany);
System.out.print("\nFor what value would you like to search? ");
int target = in.nextInt();
int index = bsearch ( myArray, target);
if (index >= 0)
{
System.out.println("The value " + target + " occurs at index " + index);
}
else
{
System.out.println("The value " + target + " does not occur in the array. ");
}
}
public static int bsearch(int[] arr, int key)
{
int lo = 0, hi = arr.length - 1;
{
while (lo < hi)
{
int mid = (lo + hi) / 2;
if (arr[mid] <= key)
lo = mid + 1;
if (arr[mid] > key)
hi = mid;
}
if (arr[lo] == key) {
return lo;
}
else if ((arr[lo] != key) && (arr[lo-1] == key)){
return lo - 1;
}
else{
System.out.print("The value " + key + " does not occur in the array. ");
}
return -1 ;
}
public static int[] getSortedRandomArray (int howMany)
{
int[] returnMe = new int [howMany];
Random rand = new Random();
for (int i = 0; i < howMany ; i++)
returnMe[i] = rand.nextInt(Integer.MAX_VALUE) + 1;
for (int i = 1; i <= (howMany - 1); i++)
{
for (int j = 0; j <= howMany - i -1; j++)
{
int tmp = 0;
if (returnMe[j] > returnMe[j+1])
{
tmp = returnMe[j];
returnMe[j] = returnMe[j + 1];
returnMe[j + 1] = tmp;
}
}
}
System.out.print("Here is a random sorted array: ");
for ( int i = 0; i < howMany; i++)
System.out.print(returnMe[i] + " ");
return returnMe;
}
最佳答案
您可以通过修改二分搜索算法代码来做到这一点,如下所示:
public static int bsearch(int[] arr, int key) {
int lo = 0, hi = arr.length - 1;
while (lo < hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= key)
lo = mid + 1;
if (arr[mid] > key)
hi = mid;
}
if (arr[lo] == key) {
return lo;
}
else {
return lo - 1;
}
}
此代码而是搜索第一个大于键的数字。可以是任何数字,6 或 10000,这并不重要。正如你所看到的,如果 arr[mid] 等于 key,代码仍然会在区间 [mid, hi] 上运行。为什么这两个最后会返回?好吧,如果输入数组与您给出的数组类似,则 lo 最终将是最后 5 个数字的索引,但如果我们在输入数组的末尾添加另一个数字,则 lo 将是最后 5 个数字后面的索引。因此,我们有 2 种不同的情况。
此外,您不能像其他答案一样使用线性循环来完成此操作,因为这会将算法减少到 O(n),并且最终只是在简化数组上进行线性搜索。
关于java - 使用二分查找查找数字的最大索引出现次数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43482720/