试题
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.)
代码
最简单的就是堆排了。距离可以提前算好,堆存index,应该还可以降低时间。
class Solution {
public int[][] kClosest(int[][] points, int K) {
PriorityQueue<int[]> queue = new PriorityQueue<int[]>(K, (int[] a, int[] b) -> euclidean(b).compareTo(euclidean(a)) );
for(int[] point : points){
queue.offer(point);
if(queue.size() > K){
queue.poll();
}
}
return queue.toArray(new int[0][0]);
}
public Double euclidean(int[] a){
int sum = 0;
for(int i = 0; i < a.length; i++){
sum += a[i] * a[i];
}
return Math.sqrt(sum);
}
}
由于题目没有要求我们返回有序的数组,只是返回K个最小数,显然如果使用快排我们的mid==K-1的话,那就找到了K个最小数。
class Solution {
int[][] points;
public int[][] kClosest(int[][] points, int K) {
this.points = points;
QuickSort(0, points.length - 1, K-1);
return Arrays.copyOfRange(points, 0, K);
}
public void QuickSort(int i, int j, int K) {
while(i < j){
int mid = partition(i, j);
if(mid == K){
return;
}else if(mid < K){
i = mid + 1;
}else if(mid > K){
j = mid - 1;
}
}
}
public int partition(int i, int j) {
int pivot = dist(i);
int left = i, right = j;
while (left < right) {
while (left < right && dist(right) >= pivot)
right--;
while (left < right && dist(left) <= pivot)
left++;
if(left < right)
swap(left, right);
}
swap(i, right);
return right;
}
public int dist(int i) {
return points[i][0] * points[i][0] + points[i][1] * points[i][1];
}
public void swap(int i, int j) {
int[] tmp = points[j];
points[j] = points[i];
points[i] = tmp;
return;
}
}