两个不同大小的数组通过索引来索引匹配比较

Question

我有两个不同大小的数组。 方案1.

A=[1,2,3] 
b=[1,3,2,4] 

场景2.

a=[1,3,2,4] 
b=[1,3,2,] 

如何都通过索引阵列匹配索引，并且如果存在任何差异显示结果。 当两个数组的长度相同时，我能够得到结果。但是当数组大小A比B短或者反之时，那么得到IndexOUTofBoundexception明显的原因是它试图访问不存在的元素。

for (int j=i;j<=i;j ++) 
if (A[i].equals(B[i])) { 
System.out.println(A[i] + "\t " + B[i]); 
} else { 
NumberOfDifference++; 
System.out.println(arrayLines1[i] + "--" + arrayLines2[i]); 
       } 

如果两个数组之间存在差异，只需要显示元素即可。方案2中的 案例Array a具有新元素4，但Array b具有空值。

Answer 1

显然你只能比较匹配索引直到最短索引。这个问题没有指定你应该用剩余的元素做什么，但是，为了便于讨论，让我们说，我们只是把它们打印：

int sharedLength = Math.min(a.length, b.length); 
for (int i = 0; i < sharedLength; ++i) { 
    if (a[i].equals(b[i])) { 
     System.out.println(a[i] + "\t" + b[i]); 
    } else { 
     System.out.println(a[i] + "--" + b[i]); 
} 

// Here, you could either use an if statement to find the longer array 
// Or, more elegantly(?), just loop over both of them. 
// One of these loops will not execute because the array is too short: 
for (int i = sharedLength; i < a.length; ++i) { 
    System.out.println(a[i] + "--"); 
} 
for (int i = sharedLength; i < b.length; ++i) { 
    System.out.println("\t--" + b[i]); 
} 

Answer 2

int m = A.size() 
int n = B.size() 

i =0 
NumberOfDifference = 0 

while (i< m && i<n) 
    if (a[i]==b[i]){   
      System.out.println(A[i] + "\t " + B[i]); 
    }else{ 

     NumberOfDifference++; 
     System.out.println(A[i] + "--" + B[i]); 
     } 
     i++; 
     } 
while (i<m) { 
      NumberOfDifference++; 
      System.out.println(A[i] + "--" + "NO Element");  
      i++ 
} 
while (i<n) { 
      NumberOfDifference++; 
      System.out.println("No Element" + "--" + B[i]);  
      i++; 
} 

Answer 3

这种方式可以让你找出不同元素：

for (int i=0; i < Math.max(A.length, B.length); i++) { 
    if (i >= A.length) { 
    NumberOfDifference++; 
    System.out.println("--" + B[i]); 
    } else if (i >= B.length) { 
    NumberOfDifference++; 
    System.out.println(A[i] + "--"); 
    } else if (A[i].equals(B[i])) { 
    System.out.println(A[i] + "\t " + B[i]); 
    } else { 
    NumberOfDifference++; 
    System.out.println(A[i] + "--" + B[i]); 
    } 
} 

这是最简单的方法：

Arrays.equals(A,B) 

Answer 4

你可以只尝试通过阵列我们遍历最短阵列的长度为最大值。

例如：

A=[1,2,3]; 
B=[1,3,2,4]; 
// Assign the shorter length to the temporary Integer minimum length 
int minimum_length = (A.length <= B.length) ? A.length : B.length 
for (int count=0; count < minimum_length; count++) { 
    if (A[count).equals(B[count]) { 
     System.out.println(A[i] + "\t " + B[i]); 
    } else { 
     NumberOfDifference++; 
     System.out.println(A[i] + "--" + B[i]); 
} 
//Now you can just add all other integers as differences to the NumberOfDifferences-Object 
for (int others = A.length; others < B.length; others++) { 
    NumberOfDifferences++; 
    System.out.println("null" + "--" + B[i]); 
} 

这应为一个选项，现在我想你可能只是转身此实现可以添加对A大于B.

长的情况下相同的选项做的工作

希望这是有用的！

Answer 5

这个问题不像其他人所说的那么清楚 所以我做出我的假设，你需要通过索引比较来做索引并显示额外的元素。

所以，你可以这样做如下：

1. 先找阵列使用数组比较小的索引和循环： 比方说，A = [1,2,3] B = [1,3 ，2,4] 所以我们循环A的长度。

   int size = A.length() < B.length() ? A.length() : B.length(); for (int i=0;i < size; i++) { if (A[i].equals(B[i])) { System.out.println(A[i] + "\t " + B[i]); } else { NumberOfDifference++; }

2. 现在你可以遍历了较大的阵列和表演的区别： 决定以同样的方式来遍历该数组： if(A.length() < B.length()) {
for(int i=size; i < A.length(); i++) { System.out.println(A[i]); } } else { for(int i=size; i < B.length(); i++) { System.out.println(B[i]); } }

Answer 6

1. 您好，您可以将用户阵列列表，但如果你坚持使用阵列 这里是我的解决方案：

   //你有两个数组： “一[]” 和 “B []”

   if (a.length == b.length){ // the code that you say have have written 
       } 
       else if (a.length > b.length) 
       { 
        int temp[] = new int[a.length]; // initialize a int[] with length of array a[] 
        for (int i = 0 ;b.length>i ; i++){ 
         temp[i] = b[i]; 
        } 
   
        for (int i = b.length; a.length> i ; i++) 
        { 
         temp[i] = -1 ; // some vaue you know wont be in your array 
        } // now you have two array with same length 
        int NumberOfDifference = 0 ; 
        for (int i=0 ; i<a.length;i++){ 
         if (a[i]==(b[i])) { 
          System.out.println("equal occured at index : " + i +" : "+ a[i] + "\t " + b[i]); 
         } 
         else 
         { 
          NumberOfDifference++; 
          System.out.println("equal occured at index : " + i +" : "+ a[i] + "\t " + b[i])); 
         } 
        } 
   
       } // a.length < b.length 
       else { 
   
        int temp[] = new int[b.length]; // initialize a int[] with length of array b[] 
        for (int i = 0 ;a.length>i ; i++){ 
         temp[i] = a[i]; 
        } 
   
        for (int i = a.length; b.length> i ; i++) 
        { 
         temp[i] = -1 ; // some vaue you know wont be in your array 
        } // now you have two array with same length 
        int NumberOfDifference = 0 ; 
        for (int i=0 ; i<b.length;i++){ 
         if (b[i]==(a[i])) { 
          System.out.println("equal occured at index : " + i +" : "+ a[i] + "\t " + b[i]); 
         } 
         else 
         { 
          NumberOfDifference++; 
          System.out.println("equal occured at index : " + i +" : "+ a[i] + "\t " + b[i])); 
         } 
        } 
   
       }