我见过其他解决方案，例如this，并想知道我使用for循环的方法是否有问题。我不想像其他人在他们的解决方案中那样使用while循环。
package com.my.practice;
public class MedianOfTwoArrays {
public static void main(String[] args) {
// Given two sorted arrays of same size
int[] a1 = {1,2,3,4,5,6};
int[] a2 = {7,8,9,10,11,12};
int[] mergedArray = new int[a1.length + a2.length];
for(int i=0 ; i < a1.length; i++){
mergedArray[i] = a1[i];
}
//System.out.println("Length:"+2*(a1.length));
for(int i= a1.length; i < 2 * (a1.length); i++) {
mergedArray[i] = a2[i];
}
for(int i=0 ; i < 2*(a1.length); i++){
System.out.println("Part of Array: "+mergedArray[i]+ " Length is: "+mergedArray.length);
}
}
}
我收到以下错误：
Length:12
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 6
at com.my.practice.MedianOfTwoArrays.main(MedianOfTwoArrays.java:30)
答案 0 :(得分：1)
依靠两个长度相同的数组并不是一个好主意，所以我不会使用a1.length * 2.此外，你不能使用相同的索引对于原始数组和合并数组，因为合并的数组更长。
建议修正：
int[] mergedArray = new int[a1.length + a2.length];
for(int i=0 ; i < a1.length; i++){
mergedArray[i] = a1[i];
}
for(int i= 0 ; i < a2.length; i++){
mergedArray[a1.length + i] = a2[i];
}
for(int i=0 ; i < mergedArray.length; i++){
System.out.println("Part of Array: "+mergedArray[i]+ " Length is: "+mergedArray.length);
}
答案 1 :(得分：1)
您的代码中的问题是由于在第二个循环中使用-r <arg>, --revision <arg>作为i的索引。它适用于a2但不适用于mergedArray
在第二个循环中使用a2作为索引
i-a1.length或使用三个指数：for(int i= a1.length; i < 2 * (a1.length); i++) {
mergedArray[i] = a2[i-a1.length];
}
，i＆amp; j。只是为了给你一个想法
k答案 2 :(得分：1)
在这一行：
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[ERROR] : download completed {"taskIdentifier":1,"data":{},"bubbles":true,"type":"downloadcompleted","source":{},"cancelBubble":false}
[ERROR] : open on /var/mobile/Containers/Data/Application/9FBAAEB4-4878-4611-8BCC-EFC112AF9777/Documents/myplvideo.mp4: File exists
[ERROR] : data : 5794757
[ERROR] : file : file:///var/mobile/Containers/Data/Application/9FBAAEB4-4878-4611-8BCC-EFC112AF9777/Documents/myplvideo.mp4
[ERROR] : file resolve path :/var/mobile/Containers/Data/Application/9FBAAEB4-4878-4611-8BCC-EFC112AF9777/Documents/myplvideo.mp4
[ERROR] : sessioncompleted {"success":true,"message":"","errorCode":0,"taskIdentifier":1,"bubbles":true,"type":"sessioncompleted","source":{},"cancelBubble":false}
您尝试从{6}到for (int i= a1.length; i < 2 * (a1.length); i++) {
mergedArray[i] = a2[i];
}
访问a2[i]。由于i是一个大小为6的数组，a2，a2[6] .. 。a2[7]不存在。
在您的情况下，您希望将值a2[11]插入a2[1]，将mergedArray[6]插入a2[2]等。
您需要在右侧减去mergedArray[7]：
a1.length或在左侧添加for (int i = a1.length; i < 2 * (a1.length); i++) {
mergedArray[i] = a2[i - a1.length];
}
：
a1.length选择更方便的一个。
答案 3 :(得分：1)
错误（对于ArrayOutOfBoundException）：
for(int i= a1.length; i < 2 * (a1.length); i++) {
mergedArray[i] = a2[i]; //Using 'i' incorrectly to access a2[i]
}
纠正上述内容：
for(int i= 0; i < a2.length; i++) {
mergedArray[i+a1.length] = a2[i]; //Using 'i' incorrectly to access a2[i]
}
原因：
答案 4 :(得分：1)
问题是你对不同长度的数组使用相同的索引：
for(int i= a1.length; i < 2 * (a1.length); i++)
mergedArray[i] = a2[i];
此处在i = 6初始，a2的数组边界为0-5，但a2[i]为a2[6]，超出范围，这会为您提供异常。
您可以使用System.arraycopy()
跳过这些循环 System.arraycopy(a1, 0, mergedArray,0, a1.length);
System.arraycopy(a2, 0, mergedArray,a1.length, a2.length);
如果仍然使用循环，只需使用另一个索引变量：
for(int i = 0, j = a1.length; i < (a2.length); i++, j++)
mergedArray[j] = a2[i];