最方便的公寓


最方便的寓所
一列street block,每个street block上都有POI (Point of Interests),比如学校,商店,诊所等等。也可能没有。

给定一个list of requirements, 比如[grocery,school],找到距离所有requirement最近的apartment位置。

follow up是如果只有一些street block有apartment怎么办

https://www.1point3acres.com/bbs ... read&tid=504803

确认一下"最远的定义"是一个点到所有最近requirement的点的总和?一个点到所有requirement中最远的距离?
sum((distance to r) for r in requirements)
or
max((distance to r) for r in requirements)
我看另一个面经的例子好像是后者

神秘文件的例子,原帖找不到了:
假设一条街上有多个block,一个block上有多个建筑。求一条街上离几个最近的特定建筑的最远距离最短的block。例子: street = [[“store”, “school”, “museum”], [“hospital”, “restaurant”], [“school”, “restaurant”], [], [“museum”]], requirement = [“store”, “museum”, “restaurant”].
第一个block到最近store是0,到最近museum是0,到最近restaurant是1,所以它的max是1
第二个block到最近store是1,到最近museum是1,到最近restaurant是0,所以它的max是1
第三个block到最近store是2,到最近museum是2,到最近restaurant是0,所以它的max是2
第四个block到最近store是3,到最近museum是1,到最近restaurant是1,所以它的max是3
第五个block到最近store是4,到最近museum是0,到最近restaurant是2,所以它的max是4
所以返回的block是第一个和第二个。


题意有误解,这题有两种问法,一种是问距离和,一种是问到最远的一个的距离,距离和的话这个作法当然是不对的,如果只是到最远的requirement 的距离,500距离0和1000都是500,这例子就没问题


粘一个个人的答案,无论是求距离和还是最远都可以用这个方法,速度为O(nk),k为requires项数。
    * Input: 1. 给一条路,路上的不同位置有不同的设施,有多个设施在不同位置的情况, List<Set<String>>
    *        2. 给一个需求设施的set
    * Output: 希望给出一个位置,距离所有设施的距离最近的和
    *
    * Example:
    *   Road: {
    *           [bookstore, school],
    *           [grocery] ,
    *           [],
    *           [],
    *           [bookstore, library],
    *           []
    *           [grocery]
    *        }
    *   Requires: [bookstore, library, grocery]
    *   Output: the best place is 4, to bookstore and lib is 0, and grocery is 2, so in sum is 2.
    * */
 
    public int findBestLocationn(List<Set<String>> road, List<String> requires) {
        Map<String, List<Integer>> roadMap = createMap(road);
        int minSum = Integer.MAX_VALUE, index = 0;
        for (int i = 0; i < road.size(); i++) {
            int sum = 0;
            for (int j = 0; j < requires.size(); j++) {
                sum += getMinLen(roadMap, requires.get(j), i);
            }
            if (sum < minSum) {
                minSum = sum;
                index = i;
            }
        }
        return index;
    }
 
    private Map<String, List<Integer>> createMap(List<Set<String>> road) {
        Map<String, List<Integer>> roadMap = new HashMap<>();
        for (int i = 0; i < road.size(); i++) {
            for (String facility: road.get(i)) {
                List<Integer> list = roadMap.getOrDefault(facility, new ArrayList<>());
                list.add(i);
                roadMap.put(facility, list);
            }
        }
        return null;
    }
 
    private int getMinLen(Map<String, List<Integer>> roadMap, String require, int index) {
        List<Integer> list = roadMap.get(require);
        int minLen = Integer.MAX_VALUE;
        for (int pos: list) {
            minLen = Math.min(minLen, Math.abs(pos-index));
        }
        return minLen;
    }



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