**[Lintcode] Max Points on a Line

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

分析：从每一个点（startPoint）出发，统计到其他点的斜率。在map中存储key=斜率，value=点数量。遇到与startPoint点重合的点，在计算所有点数量时需要加上重合点的总数。另外需要注意int除法转double时精度丢失问题。

/** * 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 {    /**     * @param points an array of point     * @return an integer     */    public int maxPoints(Point[] points) {        if(points == null) return 0;        if(points.length < 3) return points.length;                  HashMap<Double, Integer> res = new HashMap<Double, Integer>();        int max = 0;        for(int i = 0; i < points.length; i++) {            res.clear();//important            int startPoint = 1;//a            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) {                    startPoint++;                    continue;                }                int y = points[i].y - points[j].y;                int x = points[i].x - points[j].x;                double slope = 0.0;                if(x == 0) {                    slope = Integer.MAX_VALUE;                } else {                    slope = 1.0 * y / x;                  }                res.put(slope, res.containsKey(slope) ? res.get(slope) + 1 : 1);            }                        if (res.keySet().size() == 0) {                max = startPoint > max ? startPoint : max;              } else {                  for (double key : res.keySet()) {                      max = Math.max((startPoint + res.get(key)), max);                   }              }          }        return max;    }}

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