我正在处理一个面试问题,在给定非空整数数组的情况下,我需要返回k个最频繁的元素。
Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
这是我的代码:
public static void main(String[] args) {
System.out.println(topKFrequent(new int[] {1, 1, 1, 2, 2, 3, 3}, 1));
}
private static List<Integer> topKFrequent(int[] nums, int k) {
// freq map
Map<Integer, Integer> freq = new HashMap<Integer, Integer>();
for (int n : nums) {
freq.put(n, freq.getOrDefault(n, 0) + 1);
}
// bucket sort on freq
List<Integer>[] bucket = new LinkedList[nums.length + 1];
for (int i = 0; i < bucket.length; i++)
bucket[i] = new LinkedList<>();
for (int key : freq.keySet()) {
bucket[freq.get(key)].add(key);
}
// gather result
List<Integer> res = new ArrayList<>();
for (int i = bucket.length - 1; i >= 0; i--) {
res.addAll(bucket[i]);
if (res.size() >= k)
break;
}
return res;
}
我想检查一下此代码是否可以进行任何改进/优化?当存在多个具有相同频率的数字时,我感到困惑。在这种情况下我们该怎么办?
答案 0 :(得分:0)
这部分
def joinWords(string, suffix="", postfix=""):
return suffix + string + postfix
我会写一个循环
List<Integer>[] bucket = new LinkedList[nums.length + 1];
for (int i = 0; i < bucket.length; i++)
bucket[i] = new LinkedList<>();
for (int key : freq.keySet()) {
bucket[freq.get(key)].add(key);
}
在这种情况下,最后一部分
List<Integer>[] bucket = new LinkedList[nums.length + 1];
for (int key : freq.keySet()) {
List<Integer> currentBucket = bucket[freq.get(key)];
if (currentBucket == null) {
currentBucket = new LinkedList<Integer>();
bucket[freq.get(key)] = currentBucket;
}
currentBucket.add(key);
}
需要跳过空值
List<Integer> res = new ArrayList<>();
for (int i = bucket.length - 1; i >= 0; i--) {
res.addAll(bucket[i]);
if (res.size() >= k)
break;
}