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leetcode447-Number of Boomerangs

作者:互联网

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

       

 

标签:count,distance,leetcode447,map,int,Boomerangs,Number,points,Math
来源: https://www.cnblogs.com/clairexxx/p/10486905.html