Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
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public class Solution { /** * This program is used to calculate the number of points on a line on a plane. * The algorithm is simple if there is no doubt on the definition of a line. * A line can be defined using a single point and slope, one can obtain the analytic expression. * So, the algorithm in the program is the following: * 1. For a fixed point, calculate the maximum number of points on the same line with the * fixed point. * 2. run a loop with respect to each point. * * @param points -- an array of points on plane * @return maximum number of points on a line * @author Averill Zheng * @version 2014-06-01 * @since JDK 1.7 */ public
int maxPoints(Point[] points) { int
numberOfPoints = 0; int
length = points.length; for(int
i = 0; i < length; ++i){ int
max = maxPointsForFixedInterception(i, points); numberOfPoints = (numberOfPoints < max) ? max : numberOfPoints; } return
numberOfPoints; } private
int maxPointsForFixedInterception(int
i, Point[] points){ int
max = 0, length = points.length; Map<Double, Integer> slope = new
HashMap<Double, Integer>(); if(length > 0){ max = 1; if(length > 1){ //on vertical line int
number = 1; for(int
j = 0; j < length; ++j){ if(j != i){ double
x_coor = points[j].x - points[i].x; double
y_coor = points[j].y - points[i].y; //on vertical-line if(x_coor == 0
&& y_coor != 0){ ++number; max = (max < number)? number : max; } //repeat points else
if(x_coor == 0
&& y_coor == 0){ ++max; } else{ double
slopeValue = y_coor / x_coor; if(slope.containsKey(slopeValue)){ int
num = slope.get(slopeValue) + 1; max = (max < num) ? num : max; slope.put(slopeValue, num); } else{ max = (max < 2) ? 2
: max; slope.put(slopeValue, 2); } } } } } } return
max; }} |
leetcode--Max Points on a Line,布布扣,bubuko.com
leetcode--Max Points on a Line
原文地址：http://www.cnblogs.com/averillzheng/p/3765082.html
