02-线性结构4 Pop Sequence（25 分）【Java实现】

Given a stack which can keep $M$ numbers at most. Push $N$ numbers in the order of 1, 2, 3, …, $N$ and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if $M$ is 5 and $N$ is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): $M$ (the maximum capacity of the stack), $N$ (the length of push sequence), and $K$ (the number of pop sequences to be checked). Then $K$ lines follow, each contains a pop sequence of $N$ numbers. All the numbers in a line are separated by a space.

Output Specification:

For each pop sequence, print in one line “YES” if it is indeed a possible pop sequence of the stack, or “NO” if not.

Sample Input:

5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

Sample Output:

YES
NO
NO
YES
NO

---

分析

• the first line contains 3 numbers (all no more than 1000)
  三个不超过1000的数
• $M$ (the maximum capacity of the stack)
  堆栈的容量
• $N$ (the length of push sequence)
  放入的序列长度
• $K$ (the number of pop sequences to be checked)
  需要被检验的序列的数量

随机弹栈，如何判断序列是弹出的序列？

按照最初始的序列顺序，依次：
创建一个栈对象，随意放一个初始序列不包含的数（如0,-1,-2…）到栈中(避免后续判断时，存在空栈抛出异常)
遍历 结果序列：
1. 如果 当前元素>栈顶元素 且 栈容量未饱和，将初始序列的元素入栈；直到 栈容量饱和或者 栈顶元素 >= 当前元素
2. 此时判断 栈顶元素 是否和 当前元素相等，相等则符合弹栈的规律，继续判断后续元素； 否则就是虚假序列

import java.util.Scanner;
import java.util.Stack;

public class Main {

    public static boolean check(int[] arr, int m, int n){
        Stack<Integer> stack = new Stack<>();
        int stackSize = m + 1;

        int num = 1;//原始序列的元素
        stack.push(0);

        for(int i = 0; i < n; i++){//遍历序列
            //当栈未满 且 栈顶元素比 序列的当前元素小，继续入栈
            while(stack.size() < stackSize && arr[i] > stack.peek()) {
                stack.push(num);
                num++;
            }
            if(arr[i] == stack.peek()){//当栈顶元素为序列当前元素时,出栈
                stack.pop();
            }else{
                return false;
            }
        }
        return true;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        int m = in.nextInt();//堆栈的容量
        int n = in.nextInt();//序列的长度
        int k = in.nextInt();//序列的数量

        int[][] arr = new int[k][n];
        boolean[] result = new boolean[k];
        for(int i = 0; i < k; i++){
            for(int j = 0; j < n; j++){
                arr[i][j] = in.nextInt();
            }
            result[i] = check(arr[i], m, n);
        }
        in.close();

        for(int i = 0; i < result.length; i++){
            if(result[i]){
                System.out.println("YES");
            }else{
                System.out.println("NO");
            }
        }
    }
}