• 973. K Closest Points to Origin


    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. 1 <= K <= points.length <= 10000
    2. -10000 < points[i][0] < 10000
    3. -10000 < points[i][1] < 10000

    M1: min heap

    time = O(n + klogn), space = O(n)

    class Solution {
        public int[][] kClosest(int[][] points, int K) {
            int[][] res = new int[K][2];
            if(points == null || points.length == 0 || K == 0) {
                return res;
            }
            PriorityQueue<int[]> minHeap = new PriorityQueue<>(new Comparator<int[]>() {
                @Override
                public int compare(int[] p1, int[] p2) {
                    double d1 = getDistance(p1, new int[] {0, 0});
                    double d2 = getDistance(p2, new int[] {0, 0});
                    if(d1 == d2) {
                        return 0;
                    }
                    return d1 < d2 ? -1 : 1;
                }
            });
            
            for(int[] point : points) {
                minHeap.offer(point);
            }
            for(int i = 0; i < K; i++) {
                res[i] = minHeap.poll();
            }
            
            return res;
        }
        
        public double getDistance(int[] p, int[] o) {
            int x = p[0] - o[0];
            int y = p[1] - o[1];
            return Math.sqrt(x * x + y * y);
        }
    }

    M2: max heap

    time = O(k + (n-k)logk), space = O(k)

    class Solution {
        public int[][] kClosest(int[][] points, int K) {
            int[][] res = new int[K][2];
            if(points == null || points.length == 0 || K == 0) {
                return res;
            }
            PriorityQueue<int[]> maxHeap = new PriorityQueue<>(new Comparator<int[]>() {
                @Override
                public int compare(int[] p1, int[] p2) {
                    double d1 = getDistance(p1, new int[] {0, 0});
                    double d2 = getDistance(p2, new int[] {0, 0});
                    if(d1 == d2) {
                        return 0;
                    }
                    return d1 < d2 ? 1 : -1;
                }
            });
            
            for(int[] point : points) {
                if(maxHeap.size() < K) {
                    maxHeap.offer(point);
                } else if(getDistance(point, new int[] {0, 0}) < getDistance(maxHeap.peek(), new int[] {0, 0})) {
                    maxHeap.poll();
                    maxHeap.offer(point);
                }
            }
            
            for(int i = K - 1; i >= 0; i--) {
                res[i] = maxHeap.poll();
            }
            
            return res;
        }
        
        public double getDistance(int[] p, int[] o) {
            int x = p[0] - o[0];
            int y = p[1] - o[1];
            return Math.sqrt(x * x + y * y);
        }
    }

    M3: quick select

    time = O(n), space = O(logn)

    class Solution {
        int[] origin = {0, 0};
        
        public int[][] kClosest(int[][] points, int K) {
            int[][] res = new int[K][2];
            if(points == null || points.length == 0 || K == 0) {
                return res;
            }
            quickSelect(points, 0, points.length - 1, K - 1);
            for(int i = 0; i < K; i++) {
                res[i] = points[i];
            }
            return res;
        }
        
        public void quickSelect(int[][] arr, int left, int right, int target) {
            int mid = partition(arr, left, right);
            if(mid == target) {
                return;
            } else if(mid < target) {
                quickSelect(arr, mid + 1, right, target);
            } else {
                quickSelect(arr, left, mid - 1, target);
            }
        }
        
        public int partition(int[][] arr, int left, int right) {
            int[] pivot = arr[right];
            int start = left, end = right - 1;
            while(start <= end) {
                if(getDistance(arr[start], origin) < getDistance(pivot, origin)) {
                    start++;
                } else if(getDistance(arr[end], origin) >= getDistance(pivot, origin)) {
                    end--;
                } else {
                    swap(arr, start++, end--);
                }
            }
            swap(arr, start, right);
            return start;
        }
        
        public void swap(int[][] arr, int i, int j) {
            int[] tmp = arr[i];
            arr[i] = arr[j];
            arr[j] = tmp;
        }
        
        public double getDistance(int[] p, int[] o) {
            int x = p[0] - o[0];
            int y = p[1] - o[1];
            return Math.sqrt(x * x + y * y);
        }
    }
  • 相关阅读:
    C#开源爬虫NCrawler源代码解读以及将其移植到python3.2(3)
    C#开源爬虫NCrawler源代码解读以及将其移植到python3.2(2)
    C#开源爬虫NCrawler源代码解读以及将其移植到python3.2(1)
    python将项目文件打包发布的脚本(保留原来的项目结构)
    .net stream 转成 bytes[] 的实例和注意事项
    大区间素数筛选(POJ 2689)
    B. Duff in Love
    A. Duff and Meat
    NEFU_117素数个数的位数
    HDU_5504 GT and sequence
  • 原文地址:https://www.cnblogs.com/fatttcat/p/11179329.html
Copyright © 2020-2023  润新知