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Pop Sequence (25)

[编程题]Pop Sequence (25)
  • 热度指数:2185 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 64M,其他语言128M
  • 算法知识视频讲解
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.

输入描述:
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.


输出描述:
For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.
示例1

输入

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

输出

YES
NO
NO
YES
NO
推荐
本题目主要考察对栈的模拟
解题思路:
对于每行要测试的数据,单独进行模拟验证,符合要求输出YES,否则输出NO。

再验证流程:
1. 设置一个索引 index = 0,如果第一个待检测数值为X,则把 index+1 ~ X的数据全部入栈,并把 index 设置为 X,同事还要保证栈的容量不能超标。之后再弹出栈顶原素,和第一个待检测数值比较。

2. 接着判断第二待检测原素Y,如果 index > Y,则直接从栈顶弹出原素和Y比较。如果 index < Y,把 index+1 ~ Y的数据全部入栈,并把 index 设置为 Y,同事还要保证栈的容量不能超标。之后再弹出栈顶原素,和待检测数值 Y 比较。
再弹出栈顶原素和待检测数值比较时,需要判断栈顶是否为空。

3. 重复操作 2 即可

在比较过程中,发现不符号要求即可终止判断。

代码如下:
import java.io.PrintStream;
import java.util.Scanner;
import java.util.Stack;

public class Main {
	public static PrintStream out = System.out;
	public static Scanner in = new Scanner(System.in);

	public void popSequenceTest(String[] seq, int n, int capacity) {
		int[] array = new int[n];
		for(int i=0;i<n;++i){
			array[i] = Integer.valueOf(seq[i]);
		}
		
		Stack<Integer> stack = new Stack<>();
		int i,j,index = 0;
		
		for(i=0;i<n;++i){
			// 需要入栈
			if(array[i]>index){
				for(j=index+1;j<=array[i];++j){
					stack.push(j);
					// 栈溢出
					if(stack.size()>capacity){
						out.println("NO");
						return;
					}
				}
				index = array[i];
			}
			// 出栈异常
			if(stack.empty()){
				out.println("NO");
				return;
			}
			// 出栈正常
			int val = stack.pop();
			if(val != array[i]){
				// 输出的待验证的序列错误
				out.println("NO");
				return;
			}
		}
		
		out.println("YES");
	}

	public void test() {
		int M, N, K;
		M = in.nextInt();
		N = in.nextInt();
		K = in.nextInt();
		in.nextLine(); // 读取空白行

		for (int i = 1; i <= K; ++i) {
			String[] seq = in.nextLine().split(" ");
			popSequenceTest(seq, N, M);
		}
	}

	public static void main(String[] args) {
		Main m = new Main();
		m.test();
	}

}


编辑于 2015-12-13 13:39:54 回复(5)