选最合适的公寓——距离学校,超市,健身房的最大距离最小
作者:互联网
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