Max Points on a Line

最新推荐文章于 2018-07-30 23:28:01 发布

最新推荐文章于 2018-07-30 23:28:01 发布 · 68 阅读

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本文介绍了一种算法，用于在二维平面上找到最多共线的点的数量。通过使用哈希表来存储斜率和点的计数，该算法有效地解决了这一问题。对于每个点，它检查与其他所有点的关系，从而确定它们是否位于同一直线上。

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */
public class Solution {
	public int maxPoints(Point[] points) {
		if (points.length == 0 || points == null) {
			return 0;
		}
		if (points.length == 1) {
			return 1;
		}
		int res = 1;
		for (int i = 0; i < points.length; i++) {
			HashMap<Float,Integer> hashMap = new HashMap<Float, Integer>();
			int same = 0;
			int max = 1;
			for (int j = 0; j < points.length; j++) {
				if (i == j) {
					continue;
				}
				if (points[i].x == points[j].x && points[i].y == points[j].y) {
					same++;
					continue;
				}
				float tmp = (float)(points[i].y - points[j].y) / (points[i].x - points[j].x);
				if (hashMap.containsKey(tmp)) {
					hashMap.put(tmp, hashMap.get(tmp) + 1);
				} else {
					hashMap.put(tmp, 2);
				}
			}
			for (Integer value : hashMap.values()) {
				max = Math.max(max, value);
			}
			max += same;
			res = Math.max(max, res);
		}
		return res;
	}
}