问题描述
如何在不使用任何集合和Linq的情况下在C＃中实现集补和集差？
我们有两个数组：
int [] arr1 = new int { 1,2,3,4};
int[] arr2 = new int {3,4,5,6,7,8};
补码必须为：arr3 {5,8}，而差必须为：arr4 {1,2}。
我尝试过将一组添加到另一组，然后找到重复项，但未能成功。
int numDups = 0,prevIndex = 0;
for (int i = 0; i < array.Length; i++)
{
bool foundDup = false;
for (int j = 0; j < i; j++)
{
if (array[i] == array[j])
{
foundDup = true;
numDups++; // Increment means Count for Duplicate found in array.
break;
}
}
if (foundDup == false)
{
array[prevIndex] = array[i];
prevIndex++;
}
}
// Just Duplicate records replce by zero.
for (int k = 1; k <= numDups; k++)
{
array[array.Length - k] = '\0';
}
解决方法
您可以创建两个列表，一个列表用于补码，另一个列表用于差异，迭代数组A并检查B中包含哪些列表，并检查数组中不包含的列表，反之亦然，迭代B并检查A中存在哪些列表。
更新：删除列表，仅使用数组，不使用LINQ。
int[] arr1 = new int[]{ 1,2,3,4 };
int[] arr2 = new int[]{ 3,4,5,6,7,8 };
//We allocate the max possible size for each array,just for sanity
int[] arr3 = new int[arr1.Length + arr2.Length];
int[] arr4 = new int[arr1.Length + arr2.Length];
int difIndex = 0;
int compIndex = 0;
//Compute difference
foreach(int i in arr1)
{
bool found = false;
foreach(int i2 in arr2)
{
if(i == i2)
{
found = true;
break;
}
}
if(!found)
arr4[difIndex++] = i;
}
//Compute complement
foreach(int i in arr2)
{
bool found = false;
foreach(int i2 in arr1)
{
if(i == i2)
{
found = true;
break;
}
}
if(!found)
arr3[compIndex++] = i;
}
//Remove unused items from array
Array.Resize(ref arr3,compIndex);
Array.Resize(ref arr4,difIndex);
在此前提下，我们可以像这样创建函数getComplement：
int[][] getComplement(int[] arr1,int[] arr2) {
int[] complement = {};
int[] difference = {};
for (int i = 0; i < arr1.Length; i++)
{
bool isDupe = false;
for (int j = 0; j < arr2.Length; j++) {
if (arr1[i] == arr2[j] && !isDupe) {
Array.Resize(ref complement,complement.Length + 1);
complement[complement.GetUpperBound(0)] = arr2[j];
isDupe = true;
}
}
if (!isDupe) {
Array.Resize(ref difference,difference.Length + 1);
difference[difference.GetUpperBound(0)] = arr1[i];
}
}
return new[] { complement,difference };
}
，然后将其应用到我们现有的2个数组上，以获得所需的结果：
int [] arr1 = new int[] { 1,4 };
int[] arr2 = new int[] { 3,8 };
int[][] complementArr = getComplement(arr1,arr2);
int[][] differenceArr = getComplement(arr2,complementArr[0]);
int[] arr3 = differenceArr[1];
int[] arr4 = complementArr[1];
您可以看到a working demo here。
,集合A和B之间的区别是A中的所有元素都不在B中。这些集合之间的互补是B中所有不在A中的元素。这些是镜像定义，因此您实际上只需要编写一个差值方法，然后让compliment方法调用输入方法相反的差值方法即可。
（好吧，严格来说，恭维是{em> anywhere 中所有不在A中的元素，但是这里的区别是无关紧要的。）
// Get an array of all elements in b that are not in a
// This is identical to calling GetDifference with the inputs reversed so lets just do that
int[] GetCompliment(int[] a,int[] b) { return GetDifference(b,a); }
// Get an array of all elements in a that are not in b
int[] GetDifference(int[] a,int[] b)
{
// Create the buffer array at the worst-case length which is the length
// of a (where none of the elements in a are in b)
int[] c = new int[a.Length];
// Track how many elements we are actually ending up with
int length = 0;
// Loop through every element in a
foreach (var ax in a)
{
bool found = false;
// Loop through every element in b to see if it exists in a
foreach (var bx in b)
{
if (ax == bx)
{
// If the element was found in b,there's no reason to keep looping
found = true;
break;
}
}
// Only save the element if it was not found in b
if (!found)
{
c[length] = ax;
length++;
}
}
// Create the result array using the length of actual elements found
int[] result = new int[length];
// Copy the relevant slice of the buffer array into the result array
Array.Copy(c,result,length);
// Return the result array
return result;
}
用法：
int[] a = { 1,4 };
int[] b = { 3,8 };
int[] c = GetDifference(a,b);
foreach(var cx in c)
{
Console.Write(cx + ",");
}
Console.WriteLine();
int[] d = GetCompliment(a,b);
foreach(var dx in d)
{
Console.Write(dx + ",");
}
// Output:
// 1,// 5,8