[LeetCode] 149. 直线上最多的点数
题目链接 : https://leetcode-cn.com/problems/max-points-on-a-line/
题目描述:
给定一个二维平面,平面上有 n 个点,求最多有多少个点在同一条直线上。
示例:
示例 1:
输入: [[1,1],[2,2],[3,3]]
输出: 3
解释:
^
|
| o
| o
| o
+------------->
0 1 2 3 4
示例 2:
输入: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
输出: 4
解释:
^
|
| o
| o o
| o
| o o
+------------------->
0 1 2 3 4 5 6
思路:
一句话解释: 固定一点, 找其他点和这个点组成直线, 统计他们的斜率!
这里有一个重点: 求斜率.用两种方法
- 用最大约数方法(
gcd
), 把他化成最简形式,3/6 == 2/4 == 1/2
- 除数(不太精确, 速度快!)
直接写在代码里了!
class Solution:
def maxPoints(self, points: List[List[int]]) -> int:
from collections import Counter, defaultdict
# 所有点统计
points_dict = Counter(tuple(point) for point in points)
# 把唯一点列举出来
not_repeat_points = list(points_dict.keys())
n = len(not_repeat_points)
if n == 1: return points_dict[not_repeat_points[0]]
res = 0
# 求最大公约数
def gcd(x, y):
if y == 0:
return x
else:
return gcd(y, x % y)
for i in range(n - 1):
# 点1
x1, y1 = not_repeat_points[i][0], not_repeat_points[i][1]
# 斜率
slope = defaultdict(int)
for j in range(i + 1, n):
# 点2
x2, y2 = not_repeat_points[j][0], not_repeat_points[j][1]
dy, dx = y2 - y1, x2 - x1
# 方式一 利用公约数
g = gcd(dy, dx)
if g != 0:
dy //= g
dx //= g
slope["{}/{}".format(dy, dx)] += points_dict[not_repeat_points[j]]
# --------------------
# 方式二, 利用除法(不准确, 速度快)
# if dx == 0:
# tmp = "#"
# else:
# tmp = dy * 1000 / dx * 1000
# slope[tmp] += points_dict[not_repeat_points[j]]
#------------------------------
res = max(res, max(slope.values()) + points_dict[not_repeat_points[i]])
return res
java
class Solution {
public int maxPoints(int[][] points) {
int n = points.length;
if (n == 0) return 0;
if (n == 1) return 1;
int res = 0;
for (int i = 0; i < n - 1; i++) {
Map<String, Integer> slope = new HashMap<>();
int repeat = 0;
int tmp_max = 0;
for (int j = i + 1; j < n; j++) {
int dy = points[i][1] - points[j][1];
int dx = points[i][0] - points[j][0];
if (dy == 0 && dx == 0) {
repeat++;
continue;
}
int g = gcd(dy, dx);
if (g != 0) {
dy /= g;
dx /= g;
}
String tmp = String.valueOf(dy) + "/" + String.valueOf(dx);
slope.put(tmp, slope.getOrDefault(tmp, 0) + 1);
tmp_max = Math.max(tmp_max, slope.get(tmp));
}
res = Math.max(res, repeat + tmp_max + 1);
}
return res;
}
private int gcd(int dy, int dx) {
if (dx == 0) return dy;
else return gcd(dx, dy % dx);
}
}