http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1001
给出一个整数K和一个无序数组A,A的元素为N个互不相同的整数,找出数组A中所有和等于K的数对。例如K = 8,数组A:{-1,6,5,3,4,2,9,0,8},所有和等于8的数对包括(-1,9),(0,8),(2,6),(3,5)。
Input
第1行:用空格隔开的2个数,K N,N为A数组的长度。(2 <= N <= 50000,-10^9 <= K <= 10^9)
第2 - N + 1行:A数组的N个元素。(-10^9 <= A[i] <= 10^9)
Output
第1 - M行:每行2个数,要求较小的数在前面,并且这M个数对按照较小的数升序排列。
如果不存在任何一组解则输出:No Solution。
Input示例
8 9
-1
6
5
3
4
2
9
0
8
Output示例
-1 9
0 8
2 6
3 5
分析:
先排序,再用双指针,一个头一个尾,向中间扫
双指针1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57
| import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.io.StreamTokenizer; import java.util.Arrays; import java.util.LinkedHashSet;
public class Main { static class Pair { long a, b;
public Pair(long a, long b) { this.a = a; this.b = b; } }
public static void main(String[] args) throws IOException { StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); in.nextToken(); long k = (long) in.nval; in.nextToken(); int n = (int) in.nval; long[] array = new long[n]; for (int i = 0; i < n; i++) { in.nextToken(); array[i] = (long) in.nval; } Arrays.sort(array);
int front = 0, end = n - 1; LinkedHashSet<Pair> set = new LinkedHashSet<>(); while (front < end) { if (array[front] + array[end] == k) { set.add(new Pair(array[front], array[end])); front++; end--; } else if (array[front] + array[end] > k) { end--; } else { front++; } }
if (set.size() == 0) { out.println("No Solution"); } else { for (Pair pair : set) { out.println(pair.a + " " + pair.b); } } out.flush(); } }
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