标签:style blog color java os io strong for
Numeros, The Artist, had two lists A and B, such that, B was a permutation of A. Numeros was very proud of these lists. Unfortunately, while transporting them from one exhibition to another, some numbers from List A got left out. Can you find out the numbers missing from A?
Notes
Input Format
There will be four lines of input:
n - the size of the first list
This is followed by n space separated integers that make up the first list.
m - the size of the second list
This is followed by m space separated integers that make up the second list.
Output Format
Output the missing numbers in ascending order
Constraints
1<= n,m <= 1000010
1 <= x <= 10000 , x ∈ B
Xmax - Xmin < 101
题解:设置两个数组,因为x的范围在1~10000之间,只要开两个10001的数组分别记录A和B中元素的个数,然后比较两个数组就可以了。
代码如下:
1 import java.util.*; 2 3 public class Solution { 4 public static void main(String[] args) { 5 Scanner in = new Scanner(System.in); 6 int[] CountA = new int[10005]; 7 int[] CountB = new int[10005]; 8 9 int n = in.nextInt(); 10 int[] a = new int[n]; 11 for(int i = 0;i < n;i++){ 12 a[i]=in.nextInt(); 13 CountA[a[i]]++; 14 } 15 16 int m = in.nextInt(); 17 int[] b = new int[m]; 18 for(int i = 0;i < m;i++){ 19 b[i]=in.nextInt(); 20 CountB[b[i]]++; 21 } 22 23 for(int i = 1;i <= 10000;i++){ 24 if(CountB[i]>CountA[i] ) 25 System.out.printf("%d ", i); 26 } 27 System.out.println(); 28 29 30 31 } 32 }
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标签:style blog color java os io strong for
原文地址:http://www.cnblogs.com/sunshineatnoon/p/3914611.html