[抄题]:
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
Example 1:
Input: [[1,1],[2,2],[3,3]]
Output: 3
Explanation:
^
|
| o
| o
| o
+------------->
0 1 2 3 4
Example 2:
Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4
Explanation:
^
|
| o
| o o
| o
| o o
+------------------->
0 1 2 3 4 5 6
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
数组为空时,特意指出是0个点
[思维问题]:
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
分子分母同时约分掉gcd之后,用双重hashmap存储(x,(y,次数))
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 有key必然有value,如果存在x的key就不用新建value了,不存在才要建
- 每个点的duplicate也是独立的,出现在i的循环中
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O() Space complexity: O()
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
最大公约数gcd,不等于0时才能除,位置要写对
public int getGcd(int a, int b) {
//recursive
if (b == 0) return a;
//position is important
else return getGcd(b, a % b);
}
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
/**
* Definition for a point.
* class Point {
* int x;
* int y;
* Point() { x = 0; y = 0; }
* Point(int a, int b) { x = a; y = b; }
* }
*/
class Solution {
public int maxPoints(Point[] points) {
//ini
int result = 0;
//how to rewrite?
HashMap<Integer, HashMap<Integer, Integer>> map = new HashMap<>(); //cc: less than 2 points
if (points == null) return 0;
if (points.length <= 2) return points.length; //count every point
for (int i = 0; i < points.length; i++) {
int count = 0;
int duplicate = 0;
//clear previous data
map.clear(); //calculate x,y / gcd and the slope
//cc : same slope
for (int j = i + 1; j < points.length; j++) {
int x = points[i].x - points[j].x;
int y = points[i].y - points[j].y;
int gcd = getGcd(x, y);
//gcd musn't be 0
if (gcd != 0) {
x /= gcd;
y /= gcd;
}
//x stands for the difference
if (x == 0 && y == 0) {
duplicate++;
continue;
}
//map containsKey x or not, x contains y or not
if (map.containsKey(x)) {
if (map.get(x).containsKey(y)) {
map.get(x).put(y, map.get(x).get(y) + 1);
}else {
map.get(x).put(y, 1);
}
}else {
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
m.put(y, 1);
map.put(x, m);
}
count = Math.max(map.get(x).get(y), count);
}
result = Math.max(result, count + duplicate + 1);
} //return
return result;
} public int getGcd(int a, int b) {
//recursive
if (b == 0) return a;
//position is important
else return getGcd(b, a % b);
}
}