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  • leetcode447-Number of Boomerangs

    1.问题描述

    描述:

      Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between i and jequals the distance between i and k (the order of the tuple matters). Find the number of boomerangs. You may assume that n will be at most 500 and coordinates of points are all in the range [-10000, 10000] (inclusive).

    示例:

    Input:
    [[0,0],[1,0],[2,0]]
    
    Output:
    2
    
    Explanation:
    The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

    2.问题分析

    1. 第一遍:
      • 如果到点a的距离为distance的点共有n个,那么从这n个点中有序找两个点就能与A点构成回旋,根据排列公式,总的回旋个数为An2个,即n * (n - 1)个。遍历每一个点,map的key是距离,value是距离该点key个单位的点的个数。
        class Solution {
            public int numberOfBoomerangs(int[][] points) {
                int res = 0;
                for(int i = 0; i < points.length; i++){
                    Map<Double, Integer> map = new HashMap<>();
                    for(int j = 0; j < points.length; j++){
                        Double distance = Math.pow(points[i][0] - points[j][0], 2) + Math.pow(points[i][1] - points[j][1], 2);
                        if(map.containsKey(distance)){
                            map.put(distance, map.get(distance) + 1);
                        }else{
                            map.put(distance, 1);
                        }
                    }
                    
                    for(Integer count : map.values()){
                        res += count * (count - 1);
                    }
                }
                return res;
            }
        }

      • 改进:将map移到for循坏外,之后每次外部循环结束后,执行map.clear()操作。不再每次都新建map集合。
                Map<Double, Integer> map = new HashMap<>();
                for(int i = 0; i < points.length; i++){
                    for(int j = 0; j < points.length; j++){
                        Double distance = Math.pow(points[i][0] - points[j][0], 2) + Math.pow(points[i][1] - points[j][1], 2);
                        if(map.containsKey(distance)){
                            map.put(distance, map.get(distance) + 1);
                        }else{
                            map.put(distance, 1);
                        }
                    }
                    
                    for(Integer count : map.values()){
                        res += count * (count - 1);
                    }
                    map.clear();
                }
        View Code

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  • 原文地址:https://www.cnblogs.com/clairexxx/p/10486905.html
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