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692. Top K Frequent Words

2020-07-05 23:36:52  阅读:25  来源: 互联网

标签:692 love Words Top sunny queue second result words

package LeetCode_692

import java.util.*
import kotlin.collections.ArrayList
import kotlin.collections.HashMap

 * 692. Top K Frequent Words
 * https://leetcode.com/problems/top-k-frequent-words/description/
 * Given a non-empty list of words, return the k most frequent elements.
Your answer should be sorted by frequency from highest to lowest.
If two words have the same frequency, then the word with the lower alphabetical order comes first.

Example 1:
Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
Note that "i" comes before "love" due to a lower alphabetical order.

Example 2:
Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
with the number of occurrence being 4, 3, 2 and 1 respectively.

You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Input words contain only lowercase letters.

Follow up:
Try to solve it in O(n log k) time and O(n) extra space.
 * */
class Solution {
    * solution 1: HashMap+PriorityQueue, Time complexity:O(nlogk), Space complexity:O(n)
    * */
    fun topKFrequent(words: Array<String>, k: Int): List<String> {
        val map = HashMap<String, Int>()
        for (word in words) {
            map.put(word, map.getOrDefault(word, 0) + 1)
        val queue = PriorityQueue<Pair<String, Int>> { a, b ->
            if (a.second == b.second) {
                //if frequency is same, take the alphabetical desc, because the result need reverse it
            } else {
                //take the frequency asc, because the result need reverse it also
                a.second - b.second
        for (item in map) {
            queue.offer(Pair(item.key, item.value))
            if (queue.size > k) {
        val result = ArrayList<String>()
        while (queue.isNotEmpty()) {
            result.add(0, queue.poll().first)
        return result


来源: https://www.cnblogs.com/johnnyzhao/p/13252523.html


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