Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
Solution:
public int maxPoints(Point[] points) { if(points.length < 3) return points.length; int result = 2; // at least 2 points on a line Map<Double, Integer> map = new HashMap<>(); for(Point p:points) { map.clear(); int duplicate = 1; // [(1,1),(1,1),(2,2),(2,2)] -> 4 map.put(Double.MIN_VALUE, 0); // for special case: [(1,1),(1,1),(1,1)], no slope will be put to map for(Point q:points) { if(p == q) continue; if(p.x == q.x && p.y == q.y) { duplicate++; continue; } double slope = (p.x == q.x) ? Double.MAX_VALUE : (p.y-q.y)/(double)(p.x-q.x); Integer cnt = map.get(slope); if(cnt == null) cnt = 0; map.put(slope, cnt+1); } for(Integer cnt:map.values()) { result = Math.max(result, cnt+duplicate); } } return result; }
重构了下代码:
public int maxPoints(Point[] points) { if(points == null) return 0; int maxNum = 0; Map<Double, Integer> map = new HashMap<>(); for(Point p:points) { int dup = 1; map.clear(); for(Point q:points) { if(p == q) continue; if(p.x == q.x && p.y == q.y) { dup++; continue; } double slope = slope(p, q); Integer cnt = map.get(slope); if(cnt == null) cnt = 0; map.put(slope, cnt+1); } // for special case: [(1,1),(1,1),(1,1)], no slope will be put to map if(map.size() == 0) return points.length; for(int cnt:map.values()) { maxNum = Math.max(maxNum, cnt+dup); } } return maxNum; } private double slope(Point p1, Point p2) { if(p1.x == p2.x) return Double.MAX_VALUE; return (p2.y-p1.y)/((double)(p2.x-p1.x)); }