Java如何读取文本文件中的城市坐标并计算距离最近的两个城市

1. 扩展Point类 #

原有Point类缺少城市名称存储属性，先补充功能：

static class Point {
    int x, y;
    String name;
    
    public Point(int x, int y, String name) {
        this.x = x;
        this.y = y;
        this.name = name;
    }
 
    // 计算两点距离的平方，避免提前开根号损耗性能、损失精度
    public long distanceSq(Point other) {
        long dx = this.x - other.x;
        long dy = this.y - other.y;
        return dx*dx + dy*dy;
    }

    // 计算实际距离，输出时使用
    public double distance(Point other) {
        return Math.sqrt(distanceSq(other));
    }
}

2. 读取文本文件解析数据 #

使用带缓冲的BufferedReader读取文件，IO效率更高，逐行解析坐标和城市名存入列表：

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

public class ClosestCity {
    public static void main(String[] args) {
        String filePath = "你的文件路径.txt"; // 替换为实际文件路径
        List<Point> cities = new ArrayList<>();

        // 读取文件解析数据
        try (BufferedReader br = new BufferedReader(new FileReader(filePath))) {
            String line;
            while ((line = br.readLine()) != null) {
                // 按任意空白符拆分每行，过滤掉多余空格
                String[] parts = line.trim().split("\\s+");
                if (parts.length < 3) continue; // 跳过格式错误的行
                int x = Integer.parseInt(parts[0]);
                int y = Integer.parseInt(parts[1]);
                String name = parts[2];
                cities.add(new Point(x, y, name));
            }
        } catch (IOException e) {
            e.printStackTrace();
            return;
        } catch (NumberFormatException e) {
            System.out.println("坐标格式错误：" + e.getMessage());
            return;
        }

        if (cities.size() < 2) {
            System.out.println("城市数量不足2个，无法计算");
            return;
        }

        // 查找最近的两个城市
        Point cityA = cities.get(0);
        Point cityB = cities.get(1);
        long minDistSq = cityA.distanceSq(cityB);

        // 暴力枚举所有点对，数据量在10000以内时性能足够
        for (int i = 0; i < cities.size(); i++) {
            for (int j = i + 1; j < cities.size(); j++) {
                long currentSq = cities.get(i).distanceSq(cities.get(j));
                if (currentSq < minDistSq) {
                    minDistSq = currentSq;
                    cityA = cities.get(i);
                    cityB = cities.get(j);
                }
            }
        }

        // 按要求格式输出
        double dist = Math.sqrt(minDistSq);
        System.out.printf("%s and %s are closest with a distance of %.3f%n", cityA.name, cityB.name, dist);
    }
}

3. 大数据量优化 #

如果城市数量超过10000个，暴力法O(n²)的时间复杂度会性能不足，可以替换为分治算法实现的最近点对查找，时间复杂度可以降到O(nlogn)。

内容的提问来源于stack exchange，提问作者user16818633