Max Points on a Line

最新推荐文章于 2018-07-30 23:28:01 发布

原创最新推荐文章于 2018-07-30 23:28:01 发布 · 614 阅读

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本文介绍了一个算法问题：给定平面上的n个点，找出能构成一条直线的最大点集数量。通过计算各点间的斜率并进行比较，该算法能够找出共线点的最大集合。

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

/**
 * Definition for a point.
 * struct Point {
 *     int x;
 *     int y;
 *     Point() : x(0), y(0) {}
 *     Point(int a, int b) : x(a), y(b) {}
 * };
 */
class Solution {
public:
    int maxPoints(vector<Point>& points) {
        if (points.size() <= 1) {
			return points.size();
		}
		vector<double> radius;
		vector<vector<unsigned> > lines;
		vector<vector<unsigned> > duplication(points.size());
		unsigned i = 0, j;
		vector<bool> ignor(points.size(), false);
		for (; i < points.size(); i++) {
			if (ignor[i]) {
				continue;
			}
			vector<unsigned> dup;
			dup.empty();
			dup.push_back(i);
			for (j = i + 1; j < points.size(); j++) {
				double a;
				if (points[i].x == points[j].x) {
					if (points[i].y == points[j].y) {
						dup.push_back(j);
						ignor[j] = true;
						continue;
					}
					a = INT_MIN;
				} else {
					a = (double)(points[i].y - points[j].y)
							/ (double)(points[i].x - points[j].x);
				}
				if (radius.size() == 0) {
					vector<unsigned> tp;
					tp.push_back(i);
					tp.push_back(j);
					lines.push_back(tp);
					radius.push_back(a);
				} else {
					unsigned z = 0, m, n;
					bool sameline = false, has = false;
					for (; z < radius.size(); z++) {
						if (radius[z] == a) {
							for (m = 0; m < lines[z].size(); m++) {
								if (lines[z][m] == i) {
									sameline = true;
									for (n = m + 1; n < lines[z].size(); n++) {
										if (lines[z][n] == j) {
											has = true;
											break;
										}
									}
									if (!has) {
										lines[z].push_back(j);
									}
									break;
								}
							}
						}
						if (sameline)
							break;
					}
					if (!sameline) {
						vector<unsigned> tp;
						tp.push_back(i);
						tp.push_back(j);
						lines.push_back(tp);
						radius.push_back(a);
					}
				}
			}
			duplication[i]=dup;
		}
		unsigned max = 0;
		for (i = 0; i < lines.size(); i++) {
			unsigned total = 0;
			if (!ignor[lines[i][j]]) {
				total += duplication[lines[i][j]].size() - 1;
			}
			max = lines[i].size() + total > max ? lines[i].size() + total : max;
		}
		return max;
	}
};