Max Points on a Line
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
SOLUTION 1:
全部的点扫一次,然后计算每一个点与其它点之间的斜率。
创建一个MAP, KEY-VALUE是 斜率:线上的点的数目。
另外,注意重合的点,每次都要累加到每一条线上。
注意:
1. k = 0 + (double)(points[i].y - points[j].y)/(double)(points[i].x - points[j].x);
使用这个公式来计算的原因是 (double)(points[i].y - points[j].y)/(double)(points[i].x - points[j].x) 有可能计算出0和-0
0+(-0)后,就会都变成0.
2. 用Dup来计算重合的点。
3. 如果某点开始所有的点都在一起,则至少Max = Math.max(max, duplicate)。
4. 注意每次换一个点计算时,map要重建。因为即使K相同,只代表线是平等,不代表会是同一条线。
1 public class Solution { 2 public int maxPoints(Point[] points) { 3 int max = 0; 4 5 if (points == null) { 6 return 0; 7 } 8 9 int len = points.length; 10 11 for (int i = 0; i < len; i++) { 12 // Create a map to recode all the numbers of elements of every K. 13 HashMap<Double, Integer> map = new HashMap<Double, Integer>(); 14 15 // ItSelf. 16 int dup = 0; 17 18 for (int j = i; j < len; j++) { 19 // the same point. 20 if (points[i].x == points[j].x && points[i].y == points[j].y) { 21 dup++; 22 continue; 23 } 24 25 double k = Double.MAX_VALUE; 26 if (points[i].x != points[j].x) { 27 k = 0 + (double)(points[i].y - points[j].y)/(double)(points[i].x - points[j].x); 28 } 29 30 if (map.containsKey(k)) { 31 map.put(k, map.get(k) + 1); 32 } else { 33 map.put(k, 1); 34 } 35 } 36 37 max = Math.max(max, dup); 38 for (int n: map.values()) { 39 max = Math.max(max, n + dup); 40 } 41 } 42 43 return max; 44 } 45 }
主页君的GitHub:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/hash/MaxPoints.java
参考答案:
http://www.ninechapter.com/solutions/max-points-on-a-line/
时间: 2024-10-13 07:24:37