/** 方法一:哈希表 + 排序 * @param {string[]} words * @param {number} k * @return {string[]} */ var topKFrequent = function(words, k) { let count = new Map(); for(let word of words){ count.set(word, (count.get(word)||0) +1); } let res = [...count].sort((a,b)=>(b[1]-a[1])||(a[0].localeCompare(b[0]))); let rec = []; for(let i=0;i<k;i++){ rec.push(res[i][0]); } return rec; }; let words = ["i", "love", "leetcode", "i", "love", "coding"], k = 2; console.log(words, k, topKFrequent(words,k)) words= ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4 console.log(words, k, topKFrequent(words,k))
示例 1:
输入: ["i", "love", "leetcode", "i", "love", "coding"], k = 2 输出: ["i", "love"] 解析: "i" 和 "love" 为出现次数最多的两个单词,均为2次。 注意,按字母顺序 "i" 在 "love" 之前。示例 2:
输入: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4 输出: ["the", "is", "sunny", "day"] 解析: "the", "is", "sunny" 和 "day" 是出现次数最多的四个单词, 出现次数依次为 4, 3, 2 和 1 次。注意:
假定 k 总为有效值, 1 ≤ k ≤ 集合元素数。 输入的单词均由小写字母组成。