如何在原地反转数组(不创建新数组)?
问问题
28464 次
13 回答
13
public static int[] reverseArrayWithoutTempArray(int[] array) {
int i = 0, j = array.length - 1;
for (i = 0; i < array.length / 2; i++, j--) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
return array;
}
于 2012-09-04T12:53:57.353 回答
10
作业意味着仅来自我的伪代码,您通过不指定您想要的语言而变得相对容易:-)
把它变成你选择的语言:
Set i1 to index of first element in array
Set i2 to index of last element in array
while i1 < i2:
Set temporary variable to element number i1
Set element number i1 to element number i2
Set element number i2 to temporary value
Add 1 to i1
Subtract 1 from i2
理想的做法是在您的脑海中实际运行该算法,使用一张纸来跟踪变量:
- 每个数组中的元素。
i1
和i2
。temporary variable
.
对于更简单的算法,我倾向于这样做。我将调试语句插入到更难的那些中,以便计算机可以为我完成这项繁重的工作。从一张纸开始:
i1 | i2 | tempvar | el[0] | el[1] | el[2] | el[3] | el[4] | el[5]
---+----+---------+-------+-------+-------+-------+-------+------
H e l l o !
只需逐个执行这些步骤,随时检查和/或更改每一列。这将使您了解它是如何工作的,而不是仅仅给出一些代码。
于 2009-11-08T13:32:44.637 回答
3
反转字符数组而不使用 java 创建新数组。
import java.util.*;
//Reverse string array
public static void reverseArray(String[] array){
int middle = array.length / 2;
String temp;
int j = array.length -1;
for (int i = 0 ; i < middle; i++) {
temp = array[i];
array[i] = array[j];
array[j] = temp;
j--;
}
System.out.println(Arrays.toString(array));
}
如果要反转int array,则必须更改public static void reverseArray(String[] array)
aspublic static void reverseArray(int[] array)
和String temp
as int temp
。
例子:
public static void main (String[] args) throws java.lang.Exception{
String[] array = {"Smith", "Peter", "Michel", "John"};
reverseArray(array);
}
输出 :
[John, Michel, Peter, Smith]
于 2018-07-05T13:02:20.957 回答
1
不断交换末端,使用单个变量作为临时缓冲区。在伪代码中:
temp = a[0]
a[0] = a[size - 1]
a[size - 1] = temp
等等。
于 2009-11-08T13:00:13.010 回答
1
public static void main(String args[]){
int j=arr.length;
for(int i=0;i<arr.length/2;i++){
int temp=arr[i];
arr[i]=arr[j-1-i];
arr[j-1-i]=temp;}
for(int i=0;i<arr.length;i++){
System.out.println(arr[i]);
}
}
于 2016-10-10T06:42:57.057 回答
0
不要费心在内存中反转数组,只需向后迭代它!
于 2009-11-08T13:22:23.387 回答
0
这是在不使用临时变量或其他数组的情况下反转数组元素的解决方案。这仅适用于 Java 8 及更高版本。
void invertUsingStreams(Object[] arr2) {
IntStream.rangeClosed(1,arr2.length)
.mapToObj(i -> arr2[arr2.length-i])
.forEach(System.out::println);
}
谢谢,
于 2018-04-13T12:52:28.093 回答
0
var arr = [1,2,3,4,5,6]
for(let i=arr.length-2;i>=0;i--){
arr.push(arr.splice(i,1)[0])
}
console.log(arr) // [6,5,4,3,2,1]
var arr2 = [1,2,3,4,5,6]
for(let i=0,j=arr2.length-1;i<arr2.length/2;i++){
let temp = arr2[i];
arr2[i] = arr2[j-i];
arr2[j-i] = temp
}
console.log(arr2) // [6,5,4,3,2,1]
于 2021-08-08T08:50:50.257 回答
0
这是一个完整的程序,只需复制粘贴并在您的 IDE 中运行即可:
public class ReverseArrayWithoutAnotherArray {
public static void main(String[] args) {
int[] array = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int middle = array.length / 2;
int temp;
int j = array.length -1;
for(int a : array){
System.out.println(" before reverse :: " + a);
}
for (int i = 0 ; i < middle; i++, j--) {
temp = array[i];
array[i] = array[j];
array[j] = temp;
}
for(int a : array){
System.out.println(" after reverse :: " + a);
}
}
}
输出 :
before reverse :: 1
before reverse :: 2
before reverse :: 3
before reverse :: 4
before reverse :: 5
before reverse :: 6
before reverse :: 7
before reverse :: 8
before reverse :: 9
before reverse :: 10
after reverse :: 10
after reverse :: 9
after reverse :: 8
after reverse :: 7
after reverse :: 6
after reverse :: 5
after reverse :: 4
after reverse :: 3
after reverse :: 2
after reverse :: 1
于 2018-04-21T11:10:05.687 回答
0
反转字符数组而不创建新数组
public static void main(String[] args) {
char[] a = {'1', '2', '3','4'};
char temp = ' ';
for (int i = 0; i < a.length / 2; i++) {
temp = a[i];
a[i] = a[a.length - 1 - i];
a[a.length - 1 - i] = temp;
}
System.out.println(a);
}
于 2018-09-06T09:24:20.533 回答
0
- 使用临时变量临时保存变量
- 使用两个索引指针,一个来自第一个索引,最后一个来自数组长度
- 迭代数组直到数组元素的中间
- 下面是我创建的功能
- 它适用于偶数和奇数大小
- 输入数组 1 4 3 2
- 预期输出 2 3 4 1
静态 int[] reverseArray(int[] a) {
int j=a.length-1; // last index pointer
int middle = a.length/2; // end condition
for (int i=0;i<middle;i++){
int temp=a[i];
a[i]=a[j];
a[j]=temp;
j--;
}
return a;
}
于 2020-01-10T04:56:38.037 回答
0
Array.prototype.reverse = function() {
for(var i = 0, j = this.length-1; i < j; i++, j--) {
var tmp = this[i];
this[i] = this[j];
this[j] = tmp;
}
return this;
};
于 2019-04-24T21:24:43.180 回答
-1
import java.util.Scanner;
import java.util.Arrays;
public class reversearray {
private static Scanner scanner=new Scanner(System.in);
public static void main(String[] args) {
System.out.println("enter the count value ");
int n=scanner.nextInt();
int[] value=new int[n];
System.out.println("enter the elements of an array");
for(int i=0;i<n;i++)
{
value[i]=scanner.nextInt();
}
reverse(value);
}
public static void reverse(int[] array) {
int n = array.length - 1;
int temp = 0;
int count=0;
boolean swapped = true;
int mid = n / 2;
for (int i = 0; i < mid; i++) {
temp = array[i];
array[i] = array[n - i];
array[n - i] = temp;
count++;
}
if(count==(n-mid))
{
array[mid]=array[mid];
}
else
{
temp=array[mid];
array[mid]=array[mid+1];
array[mid+1]=temp;
}
System.out.println("the reveresed array elements are: " + Arrays.toString(array));
}
}
于 2020-07-27T13:37:51.563 回答