Question

我是java的新手。我想我的代码从用户获取max = 100整数的数组，然后它应该确定是否有任何重复的整数。如果有重复，则应该打印NO，如果所有整数都是唯一的则应该是YES。它应该将输入从最低到最高排序，然后在排序后打印索引号及其对应的整数。我也可以按顺序获得最终的排序输出。这就是我到目前为止所拥有的。关于如何做得更好的任何想法？

    public static void main(String args[]){            
        Scanner takeIn = new Scanner(System.in); 
        System.out.println("Enter the numbers separated by comma's");           
        String startRay = takeIn.next();           
        String[] allParts = startRay.split(",");
        int L = allParts.length;           
        int[] intAllParts = new int[allParts.length];                     
        Arrays.sort(allParts);                        
        for(int n=0; n<L; n++){             
            intAllParts[n] = Integer.parseInt(allParts[n]);               
            System.out.println(" Index of " + n + " Value of " + allParts[n]);
        }            
        for(int n=1; n<L; n++){
            intAllParts[n] = Integer.parseInt(allParts[n]);    
            if(intAllParts[n-1]==intAllParts[n]){
                   System.out.println(" First duplicate found at index " + (n));
                   System.out.println(" NO");
                   System.out.println(Arrays.toString(allParts) + " Array Values");
                   return;
            }                           
        }            
        Arrays.sort(allParts);    
        System.out.println(Arrays.toString(intAllParts) + " Array Values");
        System.out.println(" YES");           
        takeIn.close();
      }        
    }

输出

Enter the numbers separated by comma's

9,6,4,3,22,12,7,8,44,33,21,19,26,48,55,61,15,14,2,61,27,76,79,84,93

 Index of 0 Value of 12
 Index of 1 Value of 14
 Index of 2 Value of 15
 Index of 3 Value of 19
 Index of 4 Value of 2
 Index of 5 Value of 21
 Index of 6 Value of 22
 Index of 7 Value of 26
 Index of 8 Value of 27
 Index of 9 Value of 3
 Index of 10 Value of 33
 Index of 11 Value of 4
 Index of 12 Value of 44
 Index of 13 Value of 48
 Index of 14 Value of 55
 Index of 15 Value of 6
 Index of 16 Value of 61
 Index of 17 Value of 61
 Index of 18 Value of 7
 Index of 19 Value of 76
 Index of 20 Value of 79
 Index of 21 Value of 8
 Index of 22 Value of 84
 Index of 23 Value of 9
 Index of 24 Value of 93
 First duplicate found at index 17
 NO
[12, 14, 15, 19, 2, 21, 22, 26, 27, 3, 33, 4, 44, 48, 55, 6, 61, 61, 7, 76, 79, 8, 84, 9, 93] Array Values

Answer 1

只需使用TreeSet：

public static void main(String[] args) throws Exception {
    final String in = "9,6,4,3,22,12,7,8,44,33,21,19,26,48,55,61,15,14,2,61,27,76,79,84,93";
    final SortedSet<Integer> inputs = new TreeSet<>();
    boolean dup = false;
    for (final String s : in.split(",")) {
        if (!inputs.add(Integer.parseInt(s))) {
            dup = true;
        }
    }
    if (dup) {
        System.out.println("There were duplicates");
    } else {
        System.out.println("There were no duplicates");
    }
    System.out.println("The highest number is " + inputs.last());
    System.out.println("The lowest number is " + inputs.first());
    final Iterator<Integer> iter = inputs.iterator();
    for (int i = 0; iter.hasNext(); ++i) {
        System.out.println(i + " -> " + iter.next());
    }
}

Set保证唯一商品，如果商品已在add中，则false将返回Set。

SortedSet保证Set中的所有项目始终排序，TreeSet是SortedSet的实现。

首先循环输入String并按,拆分，然后将项目添加到Set并检查返回值。然后我们打印出我们是否发现重复。

然后使用last和first方法获取最大和最小的元素。

最后，使用Set循环遍历Iterator，按顺序吐出元素及其索引。请注意，Set没有索引元素，所以这些是＆＃34;伪＆＃34;指数。

Answer 2

像这样：

public static void main(String args[]){            
    Scanner takeIn = new Scanner(System.in); 
    System.out.println("Enter the numbers separated by comma's");           
    String startRay = takeIn.next();           
    String[] allParts = startRay.split(",");
    int L = allParts.length;           
    int[] intAllParts = new int[allParts.length];        
    //Arrays.sort(allParts); //no use! doesn't sort numbers, it sorts strings.
    for(int n=0; n<L; n++){             
        intAllParts[n] = Integer.parseInt(allParts[n]);               
        //System.out.println(" Index of " + n + " Value of " + allParts[n]);
        //array will be sorted, so no use printing now
    }
    Arrays.sort(intAllParts); //sorts the numbers correctly.
    for(int n=0; n<L; n++){
        //intAllParts[n] = Integer.parseInt(allParts[n]); //why do this again?
        //print now, must be intAllParts, not allParts
        System.out.println(" Index of " + n + " Value of " + intAllParts[n]);
        if(n>0 && intAllParts[n-1]==intAllParts[n]){
               System.out.println(" First duplicate found at index " + (n));
               System.out.println(" NO");
               System.out.println(Arrays.toString(intAllParts) + " Array Values");
               return;
        }                           
    }            
    //Arrays.sort(allParts);    //no use
    System.out.println(Arrays.toString(intAllParts) + " Array Values");
    System.out.println(" YES");           
    takeIn.close();
  }        
}

扫描仪输入数组排序

2 个答案: