leetcode3题解 Max Points on a line

题目大意：给出任意点（x,y）的数组，找出在同一条线上最多的点数

解题思路：直接暴力解法，三重循环，任意连接两条线，通过斜率判断其他点是否在上面

使用一个布尔变量isAllSame来判断所有的点是否一样，连接不同的两个点形成一条线，对第三个点，考虑（1）这条线如果垂直，只考虑x是否相等 （2）这条线不垂直，考虑a）是否和其他两个点一样 b）斜率是否相等

更新最大值maxCount

注意点：重复的点计算，斜率计算分母为0的情况需要特殊考虑

public class Solution {public static void main(String[] args) {Point p1 = new Point(1,1);Point p2 = new Point(1,1);Point p3 = new Point(2,3);Solution s = new Solution();Point[] points = new Point[]{p1, p2, p3};System.out.println(s.maxPoints(points));}public boolean isPointEqual(Point p1, Point p2){ //判断两个点是否相等if(p1.x == p2.x && p1.y == p2.y){return true;}return false;}public int maxPoints(Point[] points){if(points.length <= 2) return points.length;int maxCount = Integer.MIN_VALUE;boolean isAllSame = false;for(int i = 0; i < points.length; i++){for(int j = i+1; j < points.length; j++){if(!(points[i].x == points[j].x && points[i].y == points[j].y)){isAllSame = true;double ratio = Double.MAX_VALUE;int pointCount = 2;boolean isVertical = false;if(points[i].x == points[j].x){isVertical = true;}else{ratio = (double)(points[j].y - points[i].y)/(points[j].x - points[i].x);}for(int k = 0; k < points.length; k++){if(k != i && k != j){if(isVertical){if(points[k].x == points[i].x){ //连线垂直，只考虑xpointCount++;}}else{ //分别考虑点是否和其他两个点相等、斜率if(this.isPointEqual(points[k], points[i])||this.isPointEqual(points[k], points[j])){pointCount++;}else{double tmp = (double)(points[k].y - points[i].y)/(points[k].x - points[i].x);if(ratio == tmp){pointCount++;}}}}}if(pointCount > maxCount){maxCount = pointCount;}}}}if(!isAllSame){return points.length;}else{return maxCount;}}}

/** * Definition for a point. * class Point { *     int x; *     int y; *     Point() { x = 0; y = 0; } *     Point(int a, int b) { x = a; y = b; } * } */