feat: add ts solution to lc problem: No.0973

solution/0900-0999/0973.K Closest Points to Origin/README.md

@@ -71,7 +71,7 @@ tags:
 
 我们将所有点按照与原点的距离从小到大排序，然后取前 $k$ 个点即可。
 
-时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 `points` 的长度。
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 $\text{points}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -80,7 +80,7 @@ tags:
 ```python
 class Solution:
     def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
-        points.sort(key=lambda p: p[0] * p[0] + p[1] * p[1])
+        points.sort(key=lambda p: hypot(p[0], p[1]))
         return points[:k]
 ```
 
@@ -89,11 +89,8 @@ class Solution:
 ```java
 class Solution {
     public int[][] kClosest(int[][] points, int k) {
-        Arrays.sort(points, (a, b) -> {
-            int d1 = a[0] * a[0] + a[1] * a[1];
-            int d2 = b[0] * b[0] + b[1] * b[1];
-            return d1 - d2;
-        });
+        Arrays.sort(
+            points, (p1, p2) -> Math.hypot(p1[0], p1[1]) - Math.hypot(p2[0], p2[1]) > 0 ? 1 : -1);
         return Arrays.copyOfRange(points, 0, k);
     }
 }
@@ -105,8 +102,8 @@ class Solution {
 class Solution {
 public:
     vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
-        sort(points.begin(), points.end(), [](const vector<int>& a, const vector<int>& b) {
-            return a[0] * a[0] + a[1] * a[1] < b[0] * b[0] + b[1] * b[1];
+        sort(points.begin(), points.end(), [](const vector<int>& p1, const vector<int>& p2) {
+            return hypot(p1[0], p1[1]) < hypot(p2[0], p2[1]);
         });
         return vector<vector<int>>(points.begin(), points.begin() + k);
     }
@@ -118,8 +115,7 @@ public:
 ```go
 func kClosest(points [][]int, k int) [][]int {
 	sort.Slice(points, func(i, j int) bool {
-		a, b := points[i], points[j]
-		return a[0]*a[0]+a[1]*a[1] < b[0]*b[0]+b[1]*b[1]
+		return math.Hypot(float64(points[i][0]), float64(points[i][1])) < math.Hypot(float64(points[j][0]), float64(points[j][1]))
 	})
 	return points[:k]
 }
@@ -129,7 +125,8 @@ func kClosest(points [][]int, k int) [][]int {
 
 ```ts
 function kClosest(points: number[][], k: number): number[][] {
-    return points.sort((a, b) => a[0] ** 2 + a[1] ** 2 - (b[0] ** 2 + b[1] ** 2)).slice(0, k);
+    points.sort((a, b) => Math.hypot(a[0], a[1]) - Math.hypot(b[0], b[1]));
+    return points.slice(0, k);
 }
 ```
 
@@ -138,10 +135,309 @@ function kClosest(points: number[][], k: number): number[][] {
 ```rust
 impl Solution {
     pub fn k_closest(mut points: Vec<Vec<i32>>, k: i32) -> Vec<Vec<i32>> {
-        points
-            .sort_unstable_by(|a, b| (a[0].pow(2) + a[1].pow(2)).cmp(&(b[0].pow(2) + b[1].pow(2))));
-        points[0..k as usize].to_vec()
+        points.sort_by(|a, b| {
+            let dist_a = f64::hypot(a[0] as f64, a[1] as f64);
+            let dist_b = f64::hypot(b[0] as f64, b[1] as f64);
+            dist_a.partial_cmp(&dist_b).unwrap()
+        });
+        points.into_iter().take(k as usize).collect()
+    }
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- solution:start -->
+
+### 方法二：优先队列（大根堆）
+
+我们可以使用一个优先队列（大根堆）来维护距离原点最近的 $k$ 个点。
+
+时间复杂度 $O(n \times \log k)$，空间复杂度 $O(k)$。其中 $n$ 为数组 $\text{points}$ 的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
+        max_q = []
+        for i, (x, y) in enumerate(points):
+            dist = math.hypot(x, y)
+            heappush(max_q, (-dist, i))
+            if len(max_q) > k:
+                heappop(max_q)
+        return [points[i] for _, i in max_q]
+```
+
+#### Java
+
+```java
+class Solution {
+    public int[][] kClosest(int[][] points, int k) {
+        PriorityQueue<int[]> maxQ = new PriorityQueue<>((a, b) -> b[0] - a[0]);
+        for (int i = 0; i < points.length; ++i) {
+            int x = points[i][0], y = points[i][1];
+            maxQ.offer(new int[] {x * x + y * y, i});
+            if (maxQ.size() > k) {
+                maxQ.poll();
+            }
+        }
+        int[][] ans = new int[k][2];
+        for (int i = 0; i < k; ++i) {
+            ans[i] = points[maxQ.poll()[1]];
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
+        priority_queue<pair<double, int>> pq;
+        for (int i = 0, n = points.size(); i < n; ++i) {
+            double dist = hypot(points[i][0], points[i][1]);
+            pq.push({dist, i});
+            if (pq.size() > k) {
+                pq.pop();
+            }
+        }
+        vector<vector<int>> ans;
+        while (!pq.empty()) {
+            ans.push_back(points[pq.top().second]);
+            pq.pop();
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func kClosest(points [][]int, k int) [][]int {
+	maxQ := hp{}
+	for i, p := range points {
+		dist := math.Hypot(float64(p[0]), float64(p[1]))
+		heap.Push(&maxQ, pair{dist, i})
+		if len(maxQ) > k {
+			heap.Pop(&maxQ)
+		}
+	}
+	ans := make([][]int, k)
+	for i, p := range maxQ {
+		ans[i] = points[p.i]
+	}
+	return ans
+}
+
+type pair struct {
+	dist float64
+	i    int
+}
+
+type hp []pair
+
+func (h hp) Len() int { return len(h) }
+func (h hp) Less(i, j int) bool {
+	a, b := h[i], h[j]
+	return a.dist > b.dist
+}
+func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
+func (h *hp) Push(v any)   { *h = append(*h, v.(pair)) }
+func (h *hp) Pop() any     { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
+```
+
+#### TypeScript
+
+```ts
+function kClosest(points: number[][], k: number): number[][] {
+    const maxQ = new MaxPriorityQueue();
+    for (const [x, y] of points) {
+        const dist = x * x + y * y;
+        maxQ.enqueue([x, y], dist);
+        if (maxQ.size() > k) {
+            maxQ.dequeue();
+        }
+    }
+    return maxQ.toArray().map(item => item.element);
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- solution:start -->
+
+### 方法三：二分查找
+
+我们注意到，随着距离的增大，点的数量是递增的。这存在一个临界值，使得在这个值之前的点的数量小于等于 $k$，而在这个值之后的点的数量大于 $k$。
+
+因此，我们可以使用二分查找，枚举距离。每一次二分查找，我们统计出距离小于等于当前距离的点的数量，如果数量大于等于 $k$，则说明临界值在左侧，我们令右边界等于当前距离；否则，临界值在右侧，我们令左边界等于当前距禽加一。
+
+二分查找结束后，我们只需要返回距离小于等于左边界的点即可。
+
+时间复杂度 $O(n \times \log M)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\text{points}$ 的长度，而 $M$ 为距离的最大值。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
+        dist = [x * x + y * y for x, y in points]
+        l, r = 0, max(dist)
+        while l < r:
+            mid = (l + r) >> 1
+            cnt = sum(d <= mid for d in dist)
+            if cnt >= k:
+                r = mid
+            else:
+                l = mid + 1
+        return [points[i] for i, d in enumerate(dist) if d <= l]
+```
+
+#### Java
+
+```java
+class Solution {
+    public int[][] kClosest(int[][] points, int k) {
+        int n = points.length;
+        int[] dist = new int[n];
+        int r = 0;
+        for (int i = 0; i < n; ++i) {
+            int x = points[i][0], y = points[i][1];
+            dist[i] = x * x + y * y;
+            r = Math.max(r, dist[i]);
+        }
+        int l = 0;
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            int cnt = 0;
+            for (int d : dist) {
+                if (d <= mid) {
+                    ++cnt;
+                }
+            }
+            if (cnt >= k) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        int[][] ans = new int[k][0];
+        for (int i = 0, j = 0; i < n; ++i) {
+            if (dist[i] <= l) {
+                ans[j++] = points[i];
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
+        int n = points.size();
+        int dist[n];
+        int r = 0;
+        for (int i = 0; i < n; ++i) {
+            int x = points[i][0], y = points[i][1];
+            dist[i] = x * x + y * y;
+            r = max(r, dist[i]);
+        }
+        int l = 0;
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            int cnt = 0;
+            for (int d : dist) {
+                cnt += d <= mid;
+            }
+            if (cnt >= k) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        vector<vector<int>> ans;
+        for (int i = 0; i < n; ++i) {
+            if (dist[i] <= l) {
+                ans.emplace_back(points[i]);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func kClosest(points [][]int, k int) (ans [][]int) {
+	n := len(points)
+	dist := make([]int, n)
+	l, r := 0, 0
+	for i, p := range points {
+		dist[i] = p[0]*p[0] + p[1]*p[1]
+		r = max(r, dist[i])
+	}
+	for l < r {
+		mid := (l + r) >> 1
+		cnt := 0
+		for _, d := range dist {
+			if d <= mid {
+				cnt++
+			}
+		}
+		if cnt >= k {
+			r = mid
+		} else {
+			l = mid + 1
+		}
+	}
+	for i, p := range points {
+		if dist[i] <= l {
+			ans = append(ans, p)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function kClosest(points: number[][], k: number): number[][] {
+    const dist = points.map(([x, y]) => x * x + y * y);
+    let [l, r] = [0, Math.max(...dist)];
+    while (l < r) {
+        const mid = (l + r) >> 1;
+        let cnt = 0;
+        for (const d of dist) {
+            if (d <= mid) {
+                ++cnt;
+            }
+        }
+        if (cnt >= k) {
+            r = mid;
+        } else {
+            l = mid + 1;
+        }
     }
+    return points.filter((_, i) => dist[i] <= l);
 }
 ```