lintcode 996 · 最大斜率直线

题目链接： LintCode 炼码

平面上有N个点，任意2个点确定一条直线，求出所有这些直线中，斜率最大的那条直线所通过的两个点。

（点的编号为1-N，如果有多条直线斜率相等，则输出所有结果，按照点的X轴坐标排序，正序输出。数据中所有点的X轴坐标均不相等，且点坐标为随机。）

java 版本答案

public class LC996 {
    static class Point {
        int x;
        int y;

        Point() {
            x = 0;
            y = 0;
        }

        Point(int a, int b) {
            x = a;
            y = b;
        }

    }

    public static List getPoints(Point[] points) {
        /*
        思路：
        可以先将N个点按x进行排序。
        斜率k最大值为max(斜率(point[i],point[i+1])) 0<=i() {
            @Override
            public int compare(Info a, Info b) {
                //题目规定坐标x没有相等的情况。
                //因此返回值要么大于0，要么小于0
                return a.point.x-b.point.x;
            }
        });

        int[] pos = {-1,-1};
        double maxXL=-1;
        for (int i = 1; i  maxXL){
                maxXL= xl;
                pos[0]=infos[i-1].index;
                pos[1]=infos[i].index;
            }
        }

        List ans = new ArrayList<>();
        ans.add(pos[0]);
        ans.add(pos[1]);
        Collections.sort(ans);
        return ans;
    }

    static class Info{
        Point point;
        int index;
        public Info(Point p,int i){
            point =p;
            index=i;
        }
    }

    public static void main(String[] args) {
        //int[][] arr1 = {{0, 0}, {1, 2}, {2, 3}, {3, 4}};
       // int[][] arr2 = {{0, 0}, {1, 0}, {2, 0}};
        int[][] arr3 = {{-1430, -18732}, {-1441, -22482}, {-5817, 24054}, {14741, 25893}, {-24664, -28094}, {29797, 17247}, {-28277, 30109}, {13543, 18680}, {-13399, -13325}, {26225, 15630}};
        int[][] arr4 = {{13516,-22789},{-24730,22395},{21973,26520},{-23831,-13786},{-7900,1988},{11994,-16848},{24977,-2475},{8226,-22689},{-11876,-4694},{5941,-29624}};


        //Point[] p1 = new Point[arr1.length];
        //Point[] p2 = new Point[arr2.length];
        Point[] p3 = new Point[arr3.length];
        Point[] p4 = new Point[arr4.length];

//        for (int i = 0; i < arr1.length; i++) {
//            p1[i] = new Point(arr1[i][0], arr1[i][1]);
//        }
//
//        for (int i = 0; i < arr2.length; i++) {
//            p2[i] = new Point(arr2[i][0], arr2[i][1]);
//        }
//
        for (int i = 0; i < arr3.length; i++) {
            p3[i] = new Point(arr3[i][0], arr3[i][1]);
        }
        for (int i = 0; i < arr4.length; i++) {
            p4[i] = new Point(arr4[i][0], arr4[i][1]);
        }
//        System.out.println(getPoints(p1));
//        System.out.println(getPoints(p2));
        System.out.println(getPoints(p3));
        System.out.println("答案："+getPoints(p4));
    }

}

C++ 版本答案

#include
#include
#include
#include
#define ll long long int
const int INF=0x3f3f3f3f;
using namespace std;
struct head
{
    int x,y,c;    	//c用来记录是第几个点
}a[10000];
bool cmp(head x,head y)
{
    if(x.x==y.x)
        return x.y>m;
    int arr1[10][2] = {{-1430, -18732}, {-1441, -22482}, {-5817, 24054}, {14741, 25893}, {-24664, -28094}, {29797, 17247}, {-28277, 30109}, {13543, 18680}, {-13399, -13325}, {26225, 15630}};
    int arr[10][2] = {{13516,-22789},{-24730,22395},{21973,26520},{-23831,-13786},{-7900,1988},{11994,-16848},{24977,-2475},{8226,-22689},{-11876,-4694},{5941,-29624}};

    for(int i=0;i<10;i++)
    {
        a[i].x = arr[i][0];
        a[i].y= arr[i][1];
        a[i].c=i;
//        cin>>a[i].x>>a[i].y;
//        if(i!=0)
//            a[i].c=a[i-1].c+1;
    }
    sort (a,a+10,cmp);
    double max=-100000;
    for(int i=1;i<10;i++)
    {
        double n = ((double)a[i].y-a[i-1].y)/((double)a[i].x-a[i-1].x);
        cout<n)
        {
            max=max;

        }else
        {
            max=n;
            aa=a[i].c;
            bb=a[i-1].c;
        }
    }


    cout << "result is: ";
    cout<