题解 | #字符串出现次数的TopK问题#

自定义排序

public class Solution {
    /**
     * return topK string
     * @param strings string字符串一维数组 strings
     * @param k int整型 the k
     * @return string字符串二维数组
     */
    public String[][] topKstrings (String[] strings, int k) {
        // write code here
        int len = strings.length;
        String[][] res = new String[k][2];
        if (len == 0|| strings == null){
            return  res;
        }
        HashMap<String,Integer> map = new HashMap<>();
        for (int i = 0; i < len; i++) {
             String  str = strings[i];
             map.put(str,map.getOrDefault(str,0)+1);
        }
//        Arrays.sort(map,new Comparator<String,Integer>(HashMap o1) );
        ArrayList<Map.Entry<String, Integer>> list =
                new ArrayList<>(map.entrySet());
        // 先是按出现次数降序比较，相同则再按照字符ASCII码降序比较
        Collections.sort(list,
                (o1, o2) ->
                        (o1.getValue().compareTo(o2.getValue()) ==
                                0 ? o1.getKey().compareTo(o2.getKey()) :
                                o2.getValue().compareTo(o1.getValue()))
        );
        for (int i = 0; i < k; i++) {
            res[i][0] = list.get(i).getKey();
            res[i][1] = String.valueOf(list.get(i).getValue());
        }
        return res;
    }
}