단순 알고리즘 연습-제목 1004:Median

링크:http://ac.jobdu.com/problem.php?pid=1004
제목 설명:
    Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the median of the non-decreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.     Given two increasing sequences of integers, you are asked to find their median.
입력:
    Each input file may contain more than one test case.     Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (≤1000000) is the size of that sequence. Then N integers follow, separated by a space.     It is guaranteed that all the integers are in the range of long int.
출력:
    For each test case you should output the median of the two given sequences in a line.
샘플 입력:

4 11 12 13 14
5 9 10 15 16 17

샘플 출력:

13

바 쁘 지 않 을 때 예전 에 했 던 문 제 를 찾 아 연습 하 세 요.많은 알고리즘 을 다 잊 었 다!9 도 이 플랫폼 은 괜 찮 습 니 다.자신의 이전 코드 를 볼 수 있 습 니 다!
이것 은 예전 에 자바 로 만 들 었 는데 지금 은 C 를 연습 하고 있 습 니 다.

#include <stdio.h>

#define MAX 1000000
int arr1[MAX];
int arr2[MAX];
int newArr[MAX<<1];
int main(){
	int len1,len2;
	while(scanf("%d", &len1) != EOF){
		int i;
		for(i=0; i<len1; i++)
			scanf("%d",&arr1[i]);
		
		scanf("%d", &len2);
		
		for(i=0; i<len2; i++)
			scanf("%d",&arr2[i]);
		i=0;
		int index=0,j=0;
		while(i<len1 && j<len2){
			if(arr1[i] < arr2[j])
				newArr[index++] = arr1[i++];
			else
				newArr[index++] = arr2[j++];
		}

		while(i <len1)
			newArr[index++] = arr1[i++];
		while(j <len2)
			newArr[index++] = arr2[j++];

		printf("%d
",newArr[(index-1)/2]);

	}

	return 0;
}

16532 kb
10 ms
자바 코드:

import java.io.BufferedInputStream;
import java.util.Scanner;
 
public class Main {
    static int arr1[];
    static int arr2[];
    static int arr[];
    public static void main(String[] args) {
        Scanner s = new Scanner(new BufferedInputStream(System.in));
        while(s.hasNextInt()){
            int n = s.nextInt();
            arr1 = new int[n];
            for(int i=0; i<n; i++)
                arr1[i] = s.nextInt();
            int m = s.nextInt();
            arr2 = new int[m];
            arr = new int[m+n];
            for(int i=0; i<m; i++){
                arr2[i] = s.nextInt();
            }
            int i=0;
            int j=0;
            int k=0;
            while(i<n && j<m){
                if(arr1[i] <arr2[j]){
                    arr[k++] = arr1[i++];
                }
                else{
                    arr[k++] = arr2[j++];
                }
            }
            while(i<n)
                arr[k++] = arr1[i++];
            while(j<m)
                arr[k++] = arr2[j++];
            System.out.println(arr[(arr.length-1)/2]);
        }
 
    }
 
}

18368 kb
200 ms
C 와 자바 의 메모리 점용 차이 가 크 지 않 은 것 을 발 견 했 습 니 다.더 최적화 하여 배열 도 동적 으로 생 성 할 필요 가 있 습 니 다.
최적화 후:메모리 와 시간  908 kb 10 ms

#include <stdio.h>
#include <stdlib.h>
/*#define MAX 1000000*/
// int arr1[MAX];
// int arr2[MAX];
// int newArr[MAX<<1];
int * arr1,* arr2,* newArr;

int main(){
	int len1,len2;
	while(scanf("%d", &len1) != EOF){
		
		arr1 = (int *)malloc(len1*4);
		int i;
		for(i=0; i<len1; i++)
			scanf("%d",&arr1[i]);
		

		scanf("%d", &len2);
		
		arr2 = (int *)malloc(len2*4);
		newArr = (int *)malloc( (len1+len2)*4);
		for(i=0; i<len2; i++)
			scanf("%d",&arr2[i]);
		i=0;
		int index=0,j=0;
		while(i<len1 && j<len2){
			if(arr1[i] < arr2[j])
				newArr[index++] = arr1[i++];
			else
				newArr[index++] = arr2[j++];
		}

		while(i <len1)
			newArr[index++] = arr1[i++];
		while(j <len2)
			newArr[index++] = arr2[j++];

		printf("%d
",newArr[(index-1)/2]);

	}

	return 0;
}