1232. 사칙연산

링크

https://swexpertacademy.com/main/code/problem/problemDetail.do?contestProbId=AV141J8KAIcCFAYD 

SW Expert Academy

 

분류

트리

 

개인적 난이도

매우 쉬움

쉬움

보통

어려움

매우 어려움

 

핵심 알고리즘

재귀를 이용해 트리로 표현된 식을 계산한다.

 

시행착오

없음

 

코드

import java.util.*;
import java.io.*;

class Solution{
	static Node[] tree;

	public static void main(String args[]) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		for(int tc = 1; tc <= 10; ++tc){
			int N = Integer.parseInt(br.readLine());
			tree = new Node[N];

			for (int i = 0; i < N; i++){
				String[] arr = br.readLine().split(" ");

				if (arr.length == 2) tree[i] = new Node(arr[1]);
				else tree[i] = new Node(arr[1], arr[2], arr[3]);
			}

			System.out.println("#" + tc + " " + (int)tree[0].calc());
		}
	}

	public static class Node{
		double v;
		int left = -1, right = -1;

		public Node(String v, String left, String right){
			this(v);
			this.left = Integer.parseInt(left);
			this.right = Integer.parseInt(right);
		}

		public Node(String v){
			if (v.equals("+")) this.v = -1;
			else if (v.equals("-")) this.v = -2;
			else if (v.equals("*")) this.v = -3;
			else if (v.equals("/")) this.v = -4;
			else this.v = Integer.parseInt(v);
		}

		public double calc(){
			if (left == -1) return this.v;

			double c1 = tree[this.left - 1].calc(), c2 = tree[this.right - 1].calc();

			if (v == -1) return c1 + c2;
			if (v == -2) return c1 - c2;
			if (v == -3) return c1 * c2;
			return c1 / c2;
		}
	}
}