标签:sort distance int workers bucket length bikes pair
1057. Campus Bikes MediumOn a campus represented on the X-Y plane, there are n
workers and m
bikes, with n <= m
.
You are given an array workers
of length n
where workers[i] = [xi, yi]
is the position of the ith
worker. You are also given an array bikes
of length m
where bikes[j] = [xj, yj]
is the position of the jth
bike. All the given positions are unique.
Assign a bike to each worker. Among the available bikes and workers, we choose the (workeri, bikej)
pair with the shortest Manhattan distance between each other and assign the bike to that worker.
If there are multiple (workeri, bikej)
pairs with the same shortest Manhattan distance, we choose the pair with the smallest worker index. If there are multiple ways to do that, we choose the pair with the smallest bike index. Repeat this process until there are no available workers.
Return an array answer
of length n
, where answer[i]
is the index (0-indexed) of the bike that the ith
worker is assigned to.
The Manhattan distance between two points p1
and p2
is Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|
.
Example 1:
Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]] Output: [1,0] Explanation: Worker 1 grabs Bike 0 as they are closest (without ties), and Worker 0 is assigned Bike 1. So the output is [1, 0].
Example 2:
Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]] Output: [0,2,1] Explanation: Worker 0 grabs Bike 0 at first. Worker 1 and Worker 2 share the same distance to Bike 2, thus Worker 1 is assigned to Bike 2, and Worker 2 will take Bike 1. So the output is [0,2,1].
Constraints:
n == workers.length
m == bikes.length
1 <= n <= m <= 1000
workers[i].length == bikes[j].length == 2
0 <= xi, yi < 1000
0 <= xj, yj < 1000
- All worker and bike locations are unique.
class Solution { public int[] assignBikes(int[][] workers, int[][] bikes) { //calculate all the distance between workers and bikes Map<Integer,List<Pair>> map = new HashMap(); int mindis = Integer.MAX_VALUE; for(int i=0;i<workers.length;i++){ for(int j=0;j<bikes.length;j++){ int distance = Math.abs(workers[i][0]-bikes[j][0])+Math.abs(workers[i][1]-bikes[j][1]); List<Pair> list = map.getOrDefault(distance,new ArrayList()); list.add(new Pair(i,j)); map.put(distance,list); mindis = Math.min(mindis, distance); } } //get result from the bucket boolean[] status = new boolean[bikes.length]; int count = 0; int[] result = new int[workers.length]; Arrays.fill(result,-1); while(count!=result.length){ if(!map.containsKey(mindis)){ mindis++; continue; } for(Pair pair:map.get(mindis)){ if(status[pair.val]){ continue; } else{ if(result[pair.key]==-1){ status[pair.val]=true; result[pair.key]=pair.val; count++; } } } mindis++; } return result; } } class Pair{ int key; int val; Pair(int key,int val){ this.key = key; this.val = val; } }
标签:sort,distance,int,workers,bucket,length,bikes,pair 来源: https://www.cnblogs.com/cynrjy/p/16414951.html
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