PAT 1048 Find Coins 散列/雙指針

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 10​5​​ coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (≤10​5​​, the total number of coins) and M (≤10​3​​, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the two face values V​1​​ and V​2​​ (separated by a space) such that V​1​​+V​2​​=M and V​1​​≤V​2​​. If such a solution is not unique, output the one with the smallest V​1​​. If there is no solution, output 

No Solution

 instead.

Sample Input 1:

8 15
1 2 8 7 2 4 11 15
           

Sample Output 1:

4 11
           

Sample Input 2:

7 14
1 8 7 2 4 11 15
           

Sample Output 2:

No Solution
           

和LeetCode第一題相似

散列：哈希表記錄各數字出現次數，然後從小到大周遊，若i和money-i同時存在，則為最小的結果。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;

public class Main {

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

    	String[] line1=reader.readLine().split(" ");
    	int N=Integer.valueOf(line1[0]),money=Integer.valueOf(line1[1]);
    	int[] hashTable=new int[1005];//記錄各數字出現次數
    	line1=reader.readLine().split(" ");
    	for(int i=0;i<line1.length;i++) {
    		int x=Integer.valueOf(line1[i]);
    		hashTable[x]++;
    	}
    	for(int i=0;i<N;i++) {
    		if(hashTable[i]>0&&hashTable[money-i]>0) {//硬币i和money-i都有
    			if(i==money-i&&hashTable[i]<2)//币值相同，個數不能小于2
    				continue;
    			System.out.println(i+" "+(money-i));
    			return;
    		}
    	}
    	System.out.println("No Solution");
    }
}
           

雙指針：先對數組排序，然後首尾兩指針判斷兩處數字和是否等于給定值，等于則為所求答案，否則移動指針，直到i==j時得出沒有符合條件的結果。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;

public class Main {
	
    public static void main(String[] args) throws IOException {
    	BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));

    	String[] line1=reader.readLine().split(" ");
    	int N=Integer.valueOf(line1[0]),money=Integer.valueOf(line1[1]);
    	int[] num=new int[N];
    	int[] hashTable=new int[1005];//記錄各數字出現次數
    	line1=reader.readLine().split(" ");
    	for(int i=0;i<line1.length;i++) {
    		num[i]=Integer.valueOf(line1[i]);
    	}
    	Arrays.sort(num);
    	int i=0,j=N-1;//首尾雙指針
    	while(i<j) {
    		if(num[i]+num[j]==money)
    			break;
    		else if(num[i]+num[j]<money)
    			i++;
    		else
    			j--;
    	}
    	if(i==j)
    		System.out.println("No Solution");
    	else
    		System.out.println(num[i]+" "+num[j]);
    }
}