@Yano
2019-01-13T15:33:01.000000Z
字数 1548
阅读 1735
LeetCode
We have a list of points on the plane. Find the K closest points to the origin (0, 0).
(Here, the distance between two points on a plane is the Euclidean distance.)
You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.)
Example 1:
Input: points = [[1,3],[-2,2]], K = 1
Output: [[-2,2]]
Explanation:
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].
Example 2:
Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)
Note:
1 <= K <= points.length <= 10000
-10000 < points[i][0] < 10000
-10000 < points[i][1] < 10000
想到的最直接的方法,就是根据坐标的平方和排序,用一个map记录结果,key为每个点的平方和,value是结果为key的一组点。
public int[][] kClosest(int[][] points, int K) {
if (points.length < K) return null;
int[][] ans = new int[K][2];
Map<Long, List<Point>> map = new HashMap<>();
List<Long> sort = new ArrayList<>();
for (int[] point : points) {
long d = point[0] * point[0] + point[1] * point[1];
if (map.containsKey(d)) {
map.get(d).add(new Point(point));
} else {
ArrayList<Point> list = new ArrayList<Point>();
list.add(new Point(point));
map.put(d, list);
sort.add(d);
}
}
Collections.sort(sort);
int count = 0, i = 0;
while (count < K) {
long d = sort.get(i++);
List<Point> tmp = map.get(d);
for (int i1 = 0; i1 < tmp.size(); i1++) {
ans[count++] = new int[]{tmp.get(i1).getX(), tmp.get(i1).getY()};
}
}
return ans;
}
private static class Point {
private int x;
private int y;
public Point(int[] p) {
this.x = p[0];
this.y = p[1];
}
public int getX() {
return x;
}
public int getY() {
return y;
}
}