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选最合适的公寓——距离学校,超市,健身房的最大距离最小

2021-01-28 03:32:36  阅读:15  来源: 互联网

标签:blocks min int get 距离 reqs 健身房 value 最合适


refer to: https://www.algoexpert.io/questions/Apartment%20Hunting

Arrays.Apartment Hunting.Hard

  • 题目描述:

给定一组连续的block的信息链表,每个元素代表每一个街区的设备情况(是否有学校,健身房,超市),再给定一个requirements数组,包含需要考虑的元素(学校,健身房,超市)

求住在哪一个街区会使得去学校,健身房,超市的最大距离最小。

  • 分析

1. 使用三个for循环

顺序遍历每一个街区,对于每一个街区,遍历每一个reqs,计算当前的街区满足所有reqs的至少要走多少步(所有reqs满足的最大值),最大值存到一个res数组中,代表住在每一个街区的前往三个位置至少要走多少步

计算res数组的最小值,最小的(至少走多少步到达学校,超市,健身房)

  • 代码
  •  1 import java.util.*;
     2 
     3 //O(b^2*r) time | O(b) space
     4 //b is the number of blocks, r is the number of requirements
     5 class Program {
     6   public static int apartmentHunting(List<Map<String, Boolean>> blocks, String[] reqs) {
     7         int[] res = new int[blocks.size()];//the maxDistance needed to go to 3 places at each block
     8         Arrays.fill(res, Integer.MIN_VALUE);//need to calculate max value, initialize as min value
     9         
    10         for(int i = 0; i < blocks.size(); i++){//iterate all blocks
    11             for(String req: reqs){//iterate all requiements
    12                 int minReq = Integer.MAX_VALUE;//min distance to satisfy each requirement at current block
    13                 for(int j = 0; j < blocks.size(); j++){
    14                     if(blocks.get(j).get(req)){//check the list, if the boolean value is true, we find the req
    15                         minReq = Math.min(minReq, Math.abs(i-j));//update the min distance 
    16                     }                
    17                 }
    18                 res[i] = Math.max(res[i], minReq);    //update the  maxDistance needed to go to 3 places at each block        
    19             }            
    20         }
    21         return getMinIdx(res);//get the min index according to the res array.
    22         
    23   }
    24     public static int getMinIdx(int[] Array){ //function to get min_value index from an array
    25         int idx_min = 0;
    26         int min_value = Integer.MAX_VALUE;
    27         for(int i=0; i < Array.length; i++){
    28             if(min_value > Array[i]){
    29                 min_value = Array[i];
    30                 idx_min = i;
    31             }        
    32         }
    33         return idx_min;        
    34     }
    35 }

    2. 简化时间复杂度:对于每一个地点,提前计算每一个街区到达每一个地点的最短距离,存储起来,得到一个二元数组,然后对于每一个街区,求每一个街区到达三个地点至少走多少步(maxDisAtBlocks),这里得到一个一元数组存储maxDisAtBlocks,最后求maxDisAtBlocks的最小值的索引号。

代码

 1 import java.util.*;
 2 //O(br) time | O(br) space
 3 //b: # of blocks, r: # of requirements
 4 class Program {
 5   public static int apartmentHunting(List<Map<String, Boolean>> blocks, String[] reqs) {
 6     int[][] minDisFromBlocks = new int[reqs.length][];
 7         for(int i = 0; i < reqs.length; i++){
 8             minDisFromBlocks[i] = getMinDistance(blocks, reqs[i]);
 9             //get the min distance from each block to each req.
10         }
11         int[] maxDisAtBlocks = getMaxDistanceAtBlocks(blocks, minDisFromBlocks);
12         //get the max distance for each block to go to the three required places13         return getMinIdx(maxDisAtBlocks);
14     }
15     
16     public static int[] getMinDistance(List<Map<String, Boolean>> blocks, String req){
17         //the min distance from each block to each req.
18         int[] minDistance = new int[blocks.size()];
19         int closeReqIdx = Integer.MAX_VALUE;
20         for(int i = 0; i < blocks.size(); i++){
21             if(blocks.get(i).get(req)) closeReqIdx = i;
22             minDistance[i] = Math.abs(i-closeReqIdx);
23         }
24         for(int i = blocks.size() - 1; i >= 0; i--){
25             if(blocks.get(i).get(req)) closeReqIdx = i;
26             minDistance[i] = Math.min(minDistance[i], Math.abs(i-closeReqIdx));
27         }
28         return minDistance;    
29     }
30     
31     public static int[] getMaxDistanceAtBlocks(
32         //get the max distance to all reqs from each block
33     List<Map<String, Boolean>> blocks, int[][] minDisFromBlocks){
34         int[] maxDisAtBlocks = new int[blocks.size()];
35         for(int i = 0; i < blocks.size(); i++){
36             int[] minDisAtBlock = new int[minDisFromBlocks.length];
37             for(int j = 0; j < minDisFromBlocks.length; j++){
38                 minDisAtBlock[j] = minDisFromBlocks[j][i];
39             }
40             maxDisAtBlocks[i] = arrayMax(minDisAtBlock);        
41         }
42         return maxDisAtBlocks;    
43     }
44     
45     public static int getMinIdx(int[] Array){ 
46         //function to get min_value index from an array
47         int idx_min = 0;
48         int min_value = Integer.MAX_VALUE;
49         for(int i=0; i < Array.length; i++){
50             if(min_value > Array[i]){
51                 min_value = Array[i];
52                 idx_min = i;
53             }        
54         }
55         return idx_min;        
56     }
57     
58     public static int arrayMax(int[] array){
59         //get the max value from an array
60         int max = array[0];
61         for(int a: array){
62             if(a>max){
63                 max = a;
64             }
65         }
66         return max;    
67     }
68     
69 }

 

 

标签:blocks,min,int,get,距离,reqs,健身房,value,最合适
来源: https://www.cnblogs.com/LilyLiya/p/14337820.html

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