本文介绍了5最大的10个号码数组不排序的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
下面是我的code找到数字数组的最大数,但我似乎无法理解如何获得前5个号码,并将其存储在一个数组,以后再取出来。
Here's my code to find the max number in an array of numbers, but i can't seem to understand how to get the top 5 numbers and store them in an array and later retrieve them
这里的code:
public class Max {
public static void main (String[] args)
{
int i;
int large[]=new int[5];
int array[] = {33,55,13,46,87,42,10,34,43,56};
int max = array[0]; // Assume array[0] to be the max for time-being
//Looping n-1 times, O(n)
for( i = 1; i < array.length; i++) // Iterate through the First Index and compare with max
{
// O(1)
if( max < array[i])
{
// O(1)
max = array[i];// Change max if condition is True
large[i] = max;
}
}
for (int j = 0; j<5; j++)
{
System.out.println("Largest 5 : "+large[j]);
}
System.out.println("Largest is: "+ max);
// Time complexity being: O(n) * [O(1) + O(1)] = O(n)
}
}
我使用一个数组来存储5个号码,但是当我运行它,它不是我想要的。任何人都可以帮助我的计划?
I'm using an array to store 5 numbers, but when i run it, it is not what i want.Can anyone help me with the program?
推荐答案
看下面的code:
public static void main(String args[]) {
int i;
int large[] = new int[5];
int array[] = { 33, 55, 13, 46, 87, 42, 10, 34, 43, 56 };
int max = 0, index;
for (int j = 0; j < 5; j++) {
max = array[0];
index = 0;
for (i = 1; i < array.length; i++) {
if (max < array[i]) {
max = array[i];
index = i;
}
}
large[j] = max;
array[index] = Integer.MIN_VALUE;
System.out.println("Largest " + j + " : " + large[j]);
}
}
注意:如果您不想改变输入数组,然后做一个副本,做同样的操作复制的阵列上
Note: If you don't want to change the inputted array, then make a copy of it and do the same operation on the copied array.
我得到以下的输出:
最大的0:87
最大1:56
最大2:55
最大3:46
最大4:43
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