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热度指数:2192
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
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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
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:
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(); } }