具有在不同的顺序相同的元素比较阵列 [英] Comparing arrays that have same elements in different order

问题描述

我写了下面code进行比较，有相同的元素阵列但差异秩序。

I wrote below code to compare to arrays that have same elements but in diff order.

 Integer arr1[] = {1,4,6,7,2};
 Integer arr2[] = {1,2,7,4,6};

例如，以上数组相等，因为它们相同的元件1,2,4,6,7。如果你有更好的$ C $下更大的阵列，请分享。

For example, Above arrays are equal as they same elements 1,2,4,6,7. If you have better code for larger arrays, please share.

修改如果独特的元素都取自两个数组，如果他们似乎是相同的，那么也是数组应该是平等的。我怎样写的code，而无需使用任何的集合类。 例：ARR1 = {1,2,3,1,2,3} ARR2 = {3,2,1}方法应返回true（=两个数组相同）

Edit If unique elements are taken from both the arrays and if they appear to be same, then also array should be equal. How do I write the code without using any collections classes. Ex: arr1={1,2,3,1,2,3} arr2={3,2,1} Method should return true (=both arrays are same).

package com.test;

public class ArrayCompare {

public boolean compareArrays(Integer[] arr1, Integer[] arr2){
    if(arr1==null || arr2==null){
        return false;
    }
    if(arr1.length!=arr2.length){
        return false;
    }

    Integer[] sortedArr1=sortArray(arr1);
    Integer[] sortedArr2=sortArray(arr2);

    for(int i=0;i<sortedArr1.length-1;i++){
        if(sortedArr1[i]!=sortedArr2[i]){
            return false;
        }
    }
     return true;
}
public void swapElements(Integer[] arr,int pos){
    int temp=arr[pos];
    arr[pos]=arr[pos+1];
    arr[pos+1]=temp;
}
public Integer[] sortArray(Integer[] arr){
    for(int k=0;k<arr.length;k++){
        for(int i=0;i<arr.length-1;i++){
            if(arr[i]>arr[i+1]){
                swapElements(arr,i);
            }
        }
    }
    return arr;
}


public static void main(String[] args) {
    Integer arr1[] = {1,4,6,7,2};
    Integer arr2[] = {1,2,7,4,6};
    ArrayCompare arrComp=new ArrayCompare();
    System.out.println(arrComp.compareArrays(arr1, arr2));
}

}

推荐答案

你关心的重复计数？例如，你需要区分 {1，1，2} 和 {1，2，2} ？如果没有，只是用的HashSet ：

Do you care about duplicate counts? For example, would you need to distinguish between { 1, 1, 2 } and { 1, 2, 2 }? If not, just use a HashSet:

public static boolean compareArrays(Integer[] arr1, Integer[] arr2) {
    HashSet<Integer> set1 = new HashSet<Integer>(Arrays.asList(arr1));
    HashSet<Integer> set2 = new HashSet<Integer>(Arrays.asList(arr2));
    return set1.equals(set2);
}

如果您的执行的关心重复，那么无论您可以使用多集从的番石榴。

If you do care about duplicates, then either you could use a Multiset from Guava.

如果你想坚持的排序版本，为什么不使用内置的排序算法，而不是写自己的？

If you want to stick with the sorting version, why not use the built-in sorting algorithms instead of writing your own?

编辑：你甚至都不需要创建一个副本，如果你是幸福的修改现有阵列。例如：

You don't even need to create a copy, if you're happy modifying the existing arrays. For example:

public static boolean compareArrays(Integer[] arr1, Integer[] arr2) {
    Arrays.sort(arr1);
    Arrays.sort(arr2);
    return Arrays.equals(arr1, arr2);
}

您还可以有对于其中阵列是不一样的长度的情况下的优化：

You can also have an optimization for the case where the arrays aren't the same length:

public static boolean compareArrays(Integer[] arr1, Integer[] arr2) {
    // TODO: Null validation...
    if (arr1.length != arr2.length) {
        return false;
    }
    Arrays.sort(arr1);
    Arrays.sort(arr2);
    return Arrays.equals(arr1, arr2);
}

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