【Leetcode】Max Points on a Line

最新推荐文章于 2025-08-06 12:35:25 发布

yyz20002008

最新推荐文章于 2025-08-06 12:35:25 发布

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文章标签： leetcode Java Points HashMap

本文探讨了在二维平面上寻找共线点的最大数量的问题，并提供了两种解决方案：一种是通过比较点对确定共线关系；另一种是利用哈希表按斜率分类的方法来解决。两种方法都详细展示了如何计算并找出最多共线点的数量。

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

Java:

http://blog.youkuaiyun.com/linhuanmars/article/details/21060933

两种方法都和斜率有关

第一种

/**
 * 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 {
    public int maxPoints(Point[] points) {
	    if(points==null||points.length==0) return 0;
	    if(allSamePoints(points)) return points.length;
	    int max=0;
	    for(int i=0;i<points.length-1;i++)// 前后循环 限制条件一样也能过
	    {
	        for(int j=i+1;j<points.length;j++)
	        {
	            if(points[i].y==points[j].y&&points[i].x==points[j].x) continue;
	            int cur=2;
	            for(int k=0;k<points.length;k++)
	            {
	                if(k!=i&&k!=j&&(points[k].y-points[i].y)*(points[j].x-points[i].x)==(points[k].x-points[i].x)*(points[j].y-points[i].y))
	                cur++;
	                
	            }
	            max=Math.max(max,cur);
	        }
	    }
		return max;
    }
    private boolean allSamePoints(Point[] points)
    {
        for(int i=1;i<points.length;i++)
        {
            
                if(points[0].y!=points[i].y||points[0].x!=points[i].x) return false;
            
        }
        return true;
    }
}

第二种

按斜率分类用hash表

public class Solution {
    public int maxPoints(Point[] points) {
	    if(points==null||points.length==0) return 0;
	    double ratio=0.0;
	    int max=1;
	    for(int i=0;i<points.length-1;i++)
	    {
	        HashMap<Double, Integer> map= new HashMap<Double, Integer>();
	        int numofSame=0;
	        int localMax=1;
	        for(int j=i+1;j<points.length;j++)
	        {
	            if(points[j].y==points[i].y&&points[j].x==points[i].x)
	            {
	                numofSame++;
	                continue;
	            }
	            else if(points[j].x==points[i].x)
	            {
	                ratio=(double)Integer.MAX_VALUE;
	            }
	            else if(points[j].y==points[i].y)
	            {
	                ratio=0.0;
	            }
	            
	            else{
	                ratio=(double)(points[j].y-points[i].y)/(double)(points[j].x-points[i].x);
	            }
	            int num=0;
	            if(map.containsKey(ratio))
	            {
	                num=map.get(ratio)+1;
	            }
	            else
	            {
	                num=2;
	            }
	            map.put(ratio,num);
	        }
	        for( Integer value:map.values())//values!!!
	        {
	            localMax=Math.max(localMax, value);
	        }
	        localMax+=numofSame;
	        max=Math.max(localMax,max);
	    }
	    return max;
    }
}