LeetCode-Max Points on a Line[AC源码]

package com.lw.leet3;import java.util.HashMap;import java.util.Iterator;import java.util.Map;import java.util.Map.Entry;/** * @ClassName:Solution * @Description: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. * @Author LiuWei * @Date 2014年8月17日下午2:51:08 * @Mail [e-mail]  *//** * 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) {        int max = 0;        if(points.length <= 2){            max = points.length;        }        else{            for(int i = 0; i < points.length; i++){                int equalNum = 0;                Map<Double, Integer> map = new HashMap<Double, Integer>();                for(int j = i+1; j < points.length; j++ ){                    if(points[i].x == points[j].x && points[i].y == points[j].y){                        equalNum ++;                        continue;                    }                                        double k = 0;                    if(points[i].x == points[j].x){                        k = Double.MAX_VALUE;                    }                    else{                        k = ((double)(points[i].y - points[j].y)/(points[i].x - points[j].x));                        /**                         * Submission Result: Wrong Answer                         * Input:    [(2,3),(3,3),(-5,3)]                         * Output:    2                         * Expected:3                         *                          * avoid k = +0.0 -0.0                         * */                        if(k == 0){                            k = 0;                         }                    }                                        if(map.containsKey(k)){                        map.put(k, map.get(k)+1);                    }                    else{                        map.put(k, 2);                    }                }                                /**                 * Submission Result: Wrong Answer                 * Input:    [(1,1),(1,1),(1,1)]                 * Output:    0                 * Expected:3                 *                  * avoid the points are all equal                 * */                if(equalNum > max){                    max = equalNum + 1 ;                }                Iterator<Entry<Double, Integer>> iter = map.entrySet().iterator();                while(iter.hasNext()){                    Entry<Double, Integer> entry = iter.next();                    int num = entry.getValue();                    if( num + equalNum > max){                        max = num + equalNum;                     }                }            }        }        return max;    }        public static void main(String[] args){        Point[] p = {new Point(2, 3),new Point(3,3),new Point(-5,3)};//        Point[] p = {new Point(1, 1),new Point(1,1),new Point(1,1)};                Solution s = new Solution();        System.out.println(s.maxPoints(p));            }    }

自己写的时候，出了以下几点问题：
1.与x轴平行情况下，Map区分+0.0与-0.0，导致数据出错，但是不管+0.0与-0.0值是相等的，就是用+0.0 == -0.0条件判断的时候，是true。

     注：由于Map在Put数据的时候，对Key的判断是进行equal运算，而从下面的Double类的equal源码可以看到，它并不是简简单单的比较两个数据的值是否相等，而是将其调用doubleToLongBits方法，比较该方法返回的值是否相等。将这两个值输出，如下所示：
-9223372036854775808
0
由此，不难理解Map为什么区分+0.0与-0.0~

public static long doubleToLongBits(double value) {long result = doubleToRawLongBits(value);// Check for NaN based on values of bit fields, maximum// exponent and nonzero significand.if ( ((result & DoubleConsts.EXP_BIT_MASK) ==DoubleConsts. EXP_BIT_MASK) &&(result & DoubleConsts. SIGNIF_BIT_MASK) != 0L)result = 0x7ff8000000000000L;return result;}

public boolean equals(Object obj) {return (obj instanceof Double)&& ( doubleToLongBits(((Double)obj).value) ==doubleToLongBits(value));}

2.当所有的点都一样的时候，由于map的Size为0，不进入循环，但是这若干个也算作在一条直线上。所以在前面进行了处理~
总体思想：
利用平面中，点+斜率确定一条直线的思想。当固定某一个顶点的时候，如果斜率相等，那么这个直线唯一确定。
请注意下，变量equalNum