leetcode--03.位于同一条直线上点最大个数

题目：
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line
给定二维平面上的n个点，求出位于同一直线上的点的最大个数

思路：
相同的点算多个，这一点也是醉了
斜率无限大情况和相同点的情况要考虑到
还有map中的key，-0和+0算两个key
以及浮点数的精度问题

暂定这样，以后会回来修改，思路有，实现起来太恶心，还有牛客的测试用例和leetcode不一样

import java.util.Arrays;
import java.util.Map;
import java.util.HashMap;

public class Solution {
    public int maxPoints(Point[] points) {
        int length = points.length;
        if (length == 0 || length == 1 || length == 2)
            return length;
        float gradients[][] = new float[length][length];
        Map<Float, Integer> map[] = new Map[length];
        int sames[] = new int[length];
        for (int i = 0; i < length; ++i) {
            float tmp;
            map[i] = new HashMap<Float, Integer>();
            for (int j = 0; j <= i; ++j) {
                if (j == i)
                    gradients[i][j] = tmp = Float.MAX_VALUE; // 代表两个点相同
                else {
                    if (points[i].x == points[j].x) {
                        if (points[i].y == points[j].y) {
                            sames[i]++; // 代表点i前面i-1个点中有多少个相同的点
                            gradients[i][j] = tmp = Float.MAX_VALUE;
                        } else
                            gradients[i][j] = tmp = Float.MIN_VALUE; // 代表斜率为无穷
                    } else
                        gradients[i][j] = tmp = (float) (points[i].x - points[j].x)
                                / (float) (points[i].y - points[j].y);
                }
                if (tmp != Float.MAX_VALUE) {
                    Integer t = map[i].get(tmp);
                    if (t == null)
                        t = 0;
                    map[i].put(tmp, t + 1);
                }
            }
        }

        int bigest = 0;
        for (int i = 0; i < length; ++i) {
            int same = sames[i];
            if (same > bigest) {
                bigest = same;
            }
            for (int m : map[i].values()) {
                if (m == Float.MAX_VALUE)
                    continue;
                int tmp = m + same;
                if (tmp > bigest)
                    bigest = tmp;
            }
        }
        return bigest + 1; // 别忘了把此点本身加上
    }
}