LeetCode 149 — Max Points on a Line（C++ Java Python）

题目：http://oj.leetcode.com/problems/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.

题目翻译：

给定二维平面上的n个点，找出位于同一直线上的点的最大数目。
分析：

        以点为中心，计算斜率，用map(python为dictionary)保存斜率与次数的映射，注意点重合以及斜率为无穷的情况。

C++实现：

/** * Definition for a point. * struct Point { *     int x; *     int y; *     Point() : x(0), y(0) {} *     Point(int a, int b) : x(a), y(b) {} * }; */class Solution {public:    int maxPoints(vector<Point> &points) {    int maxPoints = 0;    int curPoints = 0;    int samePoints = 0;    double slope = 0;    std::map<double, int> map;    for(int i = 0; i < points.size(); ++i)    {    curPoints = 1;    samePoints = 0;    map.clear();    for(int j = i + 1; j < points.size(); ++j)    {    if(points[j].x == points[i].x && points[j].y == points[i].y){    ++samePoints;}    else    {    if(points[j].x != points[i].x)    {    slope = 1.0 * (points[j].y - points[i].y) / (points[j].x - points[i].x);    }    else    {    slope = std::numeric_limits<double>::infinity();    }    if(map.find(slope) == map.end())    {    map[slope] = 2;    }    else    {    map[slope] += 1;    }    if(map[slope] > curPoints)    {    curPoints = map[slope];    }    }    }    curPoints += samePoints;    if(curPoints > maxPoints)    {    maxPoints = curPoints;    }    }    return maxPoints;    }};

Java实现：

/** * 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 maxPoints = 0;int curPoints = 0;int samePoint = 0;double slope = 0;Map<Double, Integer> map = new HashMap<Double, Integer>();for (int i = 0; i < points.length; ++i) {curPoints = 1;samePoint = 0;map.clear();for (int j = i + 1; j < points.length; ++j) { if (points[j].x == points[i].x && points[j].y == points[i].y) {++samePoint;} else {if (points[j].x != points[i].x) {slope = 1.0 * (points[j].y - points[i].y) / (points[j].x - points[i].x) + 0.0;  // to avoid -0.0  } else {slope = Double.MAX_VALUE;}if (map.containsKey(slope)) {map.put(slope, map.get(slope) + 1);} else {map.put(slope, 2);}if (map.get(slope) > curPoints) {curPoints = map.get(slope);}}}curPoints += samePoint;if (curPoints > maxPoints) {maxPoints = curPoints;}}return maxPoints;    }}

Python实现：

# Definition for a point# class Point:#     def __init__(self, a=0, b=0):#         self.x = a#         self.y = bclass Solution:    # @param points, a list of Points    # @return an integer    def maxPoints(self, points):        if len(points) < 3:            return len(points)                maxPoints = 0                i = 0        while i < len(points):            curPoints = 1            samePoints = 0            dict = {}                        j =  i + 1;                while j < len(points):                if points[j].x == points[i].x and points[j].y == points[i].y:                    samePoints += 1                else:                    if points[j].x != points[i].x:                        slope = 1.0 * (points[j].y - points[i].y) / (points[j].x - points[i].x)                    else:                        slope = float('Inf')                                            if slope in dict:                        dict[slope] += 1                    else:                        dict[slope] = 2                                            if dict[slope] > curPoints:                        curPoints = dict[slope]                                j += 1                        curPoints += samePoints                        if curPoints > maxPoints:                maxPoints = curPoints                        i += 1                    return maxPoints

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