题目描述
给你一个数组 points ，其中 points[i] = [xi, yi] 表示 X-Y 平面上的一个点。求最多有多少个点在同一条直线上。
示例 1：
输入：points = [[1,1],[2,2],[3,3]]
输出：3
示例 2：
输入：points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
输出：4
提示：
1 <= points.length <= 300points[i].length == 2-104 <= xi, yi <= 104points 中的所有点 互不相同
解法
方法一：暴力枚举
我们可以枚举任意两个点 $(x_1, y_1), (x_2, y_2)$，把这两个点连成一条直线，那么此时这条直线上的点的个数就是 2，接下来我们再枚举其他点 $(x_3, y_3)$，判断它们是否在同一条直线上，如果在，那么直线上的点的个数就加 1，如果不在，那么直线上的点的个数不变。找出所有直线上的点的个数的最大值，即为答案。
时间复杂度 $O(n^3)$，空间复杂度 $O(1)$。其中 $n$ 是数组 points 的长度。
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16 | class Solution:
def maxPoints(self, points: List[List[int]]) -> int:
n = len(points)
ans = 1
for i in range(n):
x1, y1 = points[i]
for j in range(i + 1, n):
x2, y2 = points[j]
cnt = 2
for k in range(j + 1, n):
x3, y3 = points[k]
a = (y2 - y1) * (x3 - x1)
b = (y3 - y1) * (x2 - x1)
cnt += a == b
ans = max(ans, cnt)
return ans
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23 | class Solution {
public int maxPoints(int[][] points) {
int n = points.length;
int ans = 1;
for (int i = 0; i < n; ++i) {
int x1 = points[i][0], y1 = points[i][1];
for (int j = i + 1; j < n; ++j) {
int x2 = points[j][0], y2 = points[j][1];
int cnt = 2;
for (int k = j + 1; k < n; ++k) {
int x3 = points[k][0], y3 = points[k][1];
int a = (y2 - y1) * (x3 - x1);
int b = (y3 - y1) * (x2 - x1);
if (a == b) {
++cnt;
}
}
ans = Math.max(ans, cnt);
}
}
return ans;
}
}
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22 | class Solution {
public:
int maxPoints(vector<vector<int>>& points) {
int n = points.size();
int ans = 1;
for (int i = 0; i < n; ++i) {
int x1 = points[i][0], y1 = points[i][1];
for (int j = i + 1; j < n; ++j) {
int x2 = points[j][0], y2 = points[j][1];
int cnt = 2;
for (int k = j + 1; k < n; ++k) {
int x3 = points[k][0], y3 = points[k][1];
int a = (y2 - y1) * (x3 - x1);
int b = (y3 - y1) * (x2 - x1);
cnt += a == b;
}
ans = max(ans, cnt);
}
}
return ans;
}
};
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23 | func maxPoints(points [][]int) int {
n := len(points)
ans := 1
for i := 0; i < n; i++ {
x1, y1 := points[i][0], points[i][1]
for j := i + 1; j < n; j++ {
x2, y2 := points[j][0], points[j][1]
cnt := 2
for k := j + 1; k < n; k++ {
x3, y3 := points[k][0], points[k][1]
a := (y2 - y1) * (x3 - x1)
b := (y3 - y1) * (x2 - x1)
if a == b {
cnt++
}
}
if ans < cnt {
ans = cnt
}
}
}
return ans
}
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25 | public class Solution {
public int MaxPoints(int[][] points) {
int n = points.Length;
int ans = 1;
for (int i = 0; i < n; ++i) {
int x1 = points[i][0], y1 = points[i][1];
for (int j = i + 1; j < n; ++j) {
int x2 = points[j][0], y2 = points[j][1];
int cnt = 2;
for (int k = j + 1; k < n; ++k) {
int x3 = points[k][0], y3 = points[k][1];
int a = (y2 - y1) * (x3 - x1);
int b = (y3 - y1) * (x2 - x1);
if (a == b) {
++cnt;
}
}
if (ans < cnt) {
ans = cnt;
}
}
}
return ans;
}
}
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方法二：枚举 + 哈希表
我们可以枚举一个点 $(x_1, y_1)$，把其他所有点 $(x_2, y_2)$ 与 $(x_1, y_1)$ 连成的直线的斜率存入哈希表中，斜率相同的点在同一条直线上，哈希表的键为斜率，值为直线上的点的个数。找出哈希表中的最大值，即为答案。为了避免精度问题，我们可以将斜率 $\frac{y_2 - y_1}{x_2 - x_1}$ 进行约分，约分的方法是求最大公约数，然后分子分母同时除以最大公约数，将求得的分子分母作为哈希表的键。
时间复杂度 $O(n^2 \times \log m)$，空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别是数组 points 的长度和数组 points 所有横纵坐标差的最大值。
相似题目：
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18 | class Solution:
def maxPoints(self, points: List[List[int]]) -> int:
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
n = len(points)
ans = 1
for i in range(n):
x1, y1 = points[i]
cnt = Counter()
for j in range(i + 1, n):
x2, y2 = points[j]
dx, dy = x2 - x1, y2 - y1
g = gcd(dx, dy)
k = (dx // g, dy // g)
cnt[k] += 1
ans = max(ans, cnt[k] + 1)
return ans
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23 | class Solution {
public int maxPoints(int[][] points) {
int n = points.length;
int ans = 1;
for (int i = 0; i < n; ++i) {
int x1 = points[i][0], y1 = points[i][1];
Map<String, Integer> cnt = new HashMap<>();
for (int j = i + 1; j < n; ++j) {
int x2 = points[j][0], y2 = points[j][1];
int dx = x2 - x1, dy = y2 - y1;
int g = gcd(dx, dy);
String k = (dx / g) + "." + (dy / g);
cnt.put(k, cnt.getOrDefault(k, 0) + 1);
ans = Math.max(ans, cnt.get(k) + 1);
}
}
return ans;
}
private int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
}
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23 | class Solution {
public:
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
int maxPoints(vector<vector<int>>& points) {
int n = points.size();
int ans = 1;
for (int i = 0; i < n; ++i) {
int x1 = points[i][0], y1 = points[i][1];
unordered_map<string, int> cnt;
for (int j = i + 1; j < n; ++j) {
int x2 = points[j][0], y2 = points[j][1];
int dx = x2 - x1, dy = y2 - y1;
int g = gcd(dx, dy);
string k = to_string(dx / g) + "." + to_string(dy / g);
cnt[k]++;
ans = max(ans, cnt[k] + 1);
}
}
return ans;
}
};
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27 | func maxPoints(points [][]int) int {
n := len(points)
ans := 1
type pair struct{ x, y int }
for i := 0; i < n; i++ {
x1, y1 := points[i][0], points[i][1]
cnt := map[pair]int{}
for j := i + 1; j < n; j++ {
x2, y2 := points[j][0], points[j][1]
dx, dy := x2-x1, y2-y1
g := gcd(dx, dy)
k := pair{dx / g, dy / g}
cnt[k]++
if ans < cnt[k]+1 {
ans = cnt[k] + 1
}
}
}
return ans
}
func gcd(a, b int) int {
if b == 0 {
return a
}
return gcd(b, a%b)
}
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