Java 向堆栈添加元素

我在stack call中创建了一个方法adding在这个方法中我想在元素是特定形式后添加元素使用例如，如果堆栈中的数字是1 2 3 5，我选择数字3并输入数字4，堆栈应该是1 2 3 4 5这是我的尝试

int a[] = new int[6];
int Top = -1;

public void push() {
    if (Top > 6) {
        System.out.println(" the Stack Ovelflow");
    } else {
        Top = Top + 1;
        String m = JOptionPane.showInputDialog("enter the element stack");
        a[Top] = Integer.parseInt(m);
    }
}

public void adding() {
    String s = JOptionPane.showInputDialog("enter the element u want to add after it");
    int x = Integer.parseInt(s);
    String s2 = JOptionPane.showInputDialog("enter the element u want to add to stack");
    int d = Integer.parseInt(s2);
    for (int i = 0; i < a.length; i++) {
        if (a[i] == x) {
            a[i + 1] = d;
        }
    }
}

---

您需要确保备份数组a有足够的空间，以便插入新元素

int[] a= new int[]{1,2,3,5};  // this has only 4 elements, you can't add a 5th

所以你可以做：

public void adding(){
  // ask user for input.... and all that
  // you need an array with one more element than a. lets call it b

  int[] b = new int[a.length + 1];

  // now you need to search for x. (this is, if x is a number in your array and not an index..it wasn't clear to me)
  // so if x is a number in the array (and not the index) you need to get the index of that number:
  int index = 0;
  for (; index < a.length; index++) {  // your index variable will increment on each step
      if (a[index] == x) {
          break;            // and you break out of the loop once you found x
      }
  }

  // now you know the index of x
  // first make a copy of the partial array after x (in your example its just {5})
  int[] c = Arrays.copyOfRange(a, index, a.length);  // this will copy all elements of a from "index" to "length"

  // and here the loop that will actually insert the new number and move the rest:

   int cIndex=0;  // we need that counter later to loop through the new array c
    for (int i = 0; i < b.length; i++) { // loop through every element of b
        if (i <= index) {   // if i is currently smaller than your wanted index (there where you will find x)
            b[i] = a[i];  // then just copy the contents of a
        } else if (i == index+1) {  // we just stepped over x
            b[i] = d;         // so you can add your new number here
        } else {
            b[i] = c[cIndex];   // and here you copy the rest into b (the partial array we called c earlier)
            cIndex++;  // we need that new index, to get always the next element
        }            
    }

---

就这样。看起来很复杂，但这并不是最好或最有效的解决方案。但它是有效的，我希望它能帮助你走得更远

可能只有我一个人，但我不认为使用错误的数据结构编写代码是一个很好的教学例子。java总是有很多的集合，它比原始数组的工作更能解决这个问题。@ MattouBrou:没错，数组是最不实用的数据结构，插入中间的某个元素已经不再与栈有关。当然，我写的东西缺乏教育价值。我只是想给出一个简单的例子，说明在这种情况下如何处理数组。现在轮到OP来优化程序并切换到列表或更适合这种问题的东西。但是，在跳到列表之前先从数组开始，而不完全了解背景中发生了什么，这仍然是一件好事。也许您可以改进答案，以说明如何正确地执行此操作，或者为什么数组不适合此工作？