Check if stack elements are pairwise consecutive in Java

The

Stack is a basic data structure in computer science and is usually used for its Last-in-first-out (LIFO) attribute. When using the stack, you may encounter an interesting problem, which is to check if the elements of the stack are continuous in pairs. In this article, we will learn how to solve this problem using Java to ensure that the solution is efficient and clear.

Problem Statement

Given an integer stack, the task is to determine whether the elements of the stack are continuous in pairs. If the difference between two elements is exactly 1, they are considered continuous.

Enter

<code>4, 5, 2, 3, 10, 11</code>

Output

<code>元素是否成对连续？<br>true</code>

Step to check if stack elements are paired and continuous

The following are steps to check whether the stack elements are paired and continuous:

• Check stack size: If the number of elements on the stack is odd, the last element will not be paired, so it should be ignored for pairwise inspection.
• Pair-based check: Loop through the stack, pop the elements in pairs, and check if they are continuous.
• Restore stack: After performing a check, the stack should be restored to its original state.

Java program used to check whether stack elements are in pairs and contiguous

The following is a program in Java that checks whether stack elements are in pairs and contiguous:

import java.util.Stack;

public class PairwiseConsecutiveChecker {
    public static boolean areElementsPairwiseConsecutive(Stack<Integer> stack) {
        // 基本情况：如果堆栈为空或只有一个元素，则返回 true
        if (stack.isEmpty() || stack.size() == 1) {
            return true;
        }

        // 使用临时堆栈在检查时保存元素
        Stack<Integer> tempStack = new Stack<>();
        boolean isPairwiseConsecutive = true;

        // 成对处理堆栈元素
        while (!stack.isEmpty()) {
            int first = stack.pop();
            tempStack.push(first);

            if (!stack.isEmpty()) {
                int second = stack.pop();
                tempStack.push(second);

                // 检查这对元素是否连续
                if (Math.abs(first - second) != 1) {
                    isPairwiseConsecutive = false;
                }
            }
        }

        // 恢复原始堆栈
        while (!tempStack.isEmpty()) {
            stack.push(tempStack.pop());
        }

        return isPairwiseConsecutive;
    }

    public static void main(String[] args) {
        Stack<Integer> stack = new Stack<>();
        stack.push(4);
        stack.push(5);
        stack.push(2);
        stack.push(3);
        stack.push(10);
        stack.push(11);

        boolean result = areElementsPairwiseConsecutive(stack);

        System.out.println("元素是否成对连续？ " + result);
    }
}

Explanation

Restore stack: Since we modified the stack when checking the right, it is very important to restore it to its original state after the check is completed. This ensures that the stack remains unchanged for any subsequent operations.

Edge case: This function handles edge cases, such as an empty stack or a stack with only one element, returning true because these cases satisfy the condition insignificantly.

Time Complexity: The time complexity of this method is O(n), where n is the number of elements in the stack. This is because we only traverse the stack once, popping and pressing elements as needed.

Space complexity: Because the temporary stack is used, the space complexity is also O(n).

Conclusion

This solution provides an efficient way to check if elements in the stack are continuous in pairs. The key is to process the stack in pairs and make sure the stack is restored to its original state after the operation. This approach maintains the integrity of the stack while providing a clear and effective solution.

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