05 August 2008
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:
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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:
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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:
计算哪个点离原点最近，可以利用众所周知的距离公式来计算，但是大可不必用外面的那一层根号，因为仅仅内部平方和就可以用来判断了。
实则是可以转化成了topK问题，所以topK问题的解法都可以用，
1）全排序， time: O(n) = N log(N)，space: O(N)。
2）堆排序，用maxHeap，大顶堆堆顶元素最大，因为我们要找的是smallest one，我们需要不断判断堆顶的元素与新元素的大小关系。 先加入k个元素（计算后的）到maxheap中。 进行逻辑判断，堆顶和新元素，实施必要的替换，保证heap的size。 简单循环，取出heap中的值到array中，输出。 time: O(n) = Nlog(k)，space: O(k)。
3）BFPRT算法，快排的思想，time: O(n) = N，space: O(N)。
优先掌握，快选模板，将k个原点一次选选到pivot - k的左边，最坏O(n^2)，最优O(n)，空间复杂度O(1)
将快选模板记住
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class Solution {
public int[][] kClosest(int[][] points, int k) {
return quickSelect(points, k, 0, points.length - 1);
}
private int[][] quickSelect(int[][] points, int k, int left, int right) {
while (left < right) {
int mid = partition(points, left, right);
if(mid == k) {
break;
} else if (mid < k) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return Arrays.copyOf(points, k);
}
private int partition(int[][] points, int left, int right) {
int[] pivot = points[left]; // 快选模板，用left为pivot
int i = left + 1;
int j = right;
while (i <= j) {
while (i <= j && cal(points[i]) <= cal(pivot)) {
i++;
}
while (i <= j && cal(points[j]) > cal(pivot)) {
j--;
}
if (i > j) {
break;
}
swap(points, i, j);
}
swap(points, left, j);
return j;
}
private int cal(int[] a) {
return a[0] * a[0] + a[1] * a[1];
}
private void swap(int[][] points, int i, int j) {
int[] temp = points[i];
points[i] = points[j];
points[j] = temp;
}
}
全排序
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class Solution {
public int[][] kClosest(int[][] points, int K) {
int[][] ans=new int[K][2];
Arrays.sort(points,(int[] o1,int[] o2)->(o1[0]*o1[0]+o1[1]*o1[1]-o2[0]*o2[0]-o2[1]*o2[1]));
System.arraycopy(points,0,ans,0,K);
return ans;
}
}
堆排序
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class Solution {
public int[][] kClosest(int[][] points, int K) {
PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> (cal(b) - cal(a))); // 倒序排列
// Queue<int[]> priorityQueue = new PriorityQueue<>(K, (o1, o2) -> o1[0] * o1[0] + o1[1] * o1[1] - (o2[0] * o2[0] + o2[1] * o2[1]));
int[][] result = new int[K][2];
for (int i = 0; i < K; i++) { // 先加入K个元素
maxHeap.offer(points[i]);
}
for (int i = K; i < points.length; i++) {
if (cal(maxHeap.peek()) > cal(points[i])) { // 堆顶最大距离要比新来的点大，移出
maxHeap.poll();
maxHeap.offer(points[i]);
}
}
for (int i = 0; i < K; i++) {
result[i] = maxHeap.poll();
}
return result;
}
private int cal(int[] a) { // 函数编程计算坐标距离
return a[0] * a[0] + a[1] * a[1];
}
}
同样堆排序，略微不同的实现
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class Solution {
public int[][] kClosest(int[][] points, int K) {
if(points.length == 0 || points[0].length == 0) return points;
PriorityQueue<int[]> pq = new PriorityQueue<>(
(a, b) -> ((a[0]*a[0] + a[1]*a[1]) - (b[0]*b[0] + b[1]*b[1])));
for (int i = 0; i < points.length; i++) { // 先加入所有点
pq.add(points[i]);
}
int[][] res = new int[K][2];
for (int i = 0; i < K; i++) { // 然后取出K个点
int[] temp = pq.poll();
res[i][0] = temp[0];
res[i][1] = temp[1];
}
return res;
}
}
pq不使用lamda
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class Solution {
static class Pair implements Comparable<Pair>{
int first;
int second[];
Pair(int first, int second[]){
this.first = first;
this.second = second;
}
public int compareTo(Pair o){
return this.first - o.first;
}
}
public int[][] kClosest(int[][] arr, int k) {
int n = arr.length;
int dist[] = new int[n];
PriorityQueue<Pair> pq = new PriorityQueue<>(Collections.reverseOrder());
// calculate distance between them
for(int i=0; i<n; i++)
dist[i] = arr[i][0]*arr[i][0] + arr[i][1]*arr[i][1];
for(int i=0; i<arr.length; i++){
pq.add(new Pair(dist[i], arr[i] ));
if(pq.size()>k)
pq.remove();
}
int ans[][]= new int[k][2];
int i=0;
while(!pq.isEmpty()){
ans[i] = pq.remove().second;
i++;
}
return ans;
}
}