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给一非空的单词列表,返回前 k 个出现次数最多的单词。 返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率,按字母顺序排序。 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。
给一非空的单词列表,返回前 k 个出现次数最多的单词。
返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率,按字母顺序排序。
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。
用一个map来保存各个单词出现的次数,然后取出它们的key值组成数组,对数组按元素个数进行排序,当元素个数相同时,返回排在前面的字母
/** * @param {string[]} words * @param {number} k * @return {string[]} */ var topKFrequent = function(words, k) { const map = new Map(); for (let i = 0, length = words.length; i < length; i++) { const current = words[i]; if (map.get(current)) { map.set(current, map.get(current) + 1) } else { map.set(current, 1); } } const keyArray = [...map.keys()]; console.log(keyArray); keyArray.sort((a, b) => { let numberOfA = map.get(a); let numberOfB = map.get(b); if (numberOfA === numberOfB) { return a > b ? 1 : -1; } return numberOfB - numberOfA; }); return keyArray.slice(0, k); };
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题目
思路
用一个map来保存各个单词出现的次数,然后取出它们的key值组成数组,对数组按元素个数进行排序,当元素个数相同时,返回排在前面的字母
解答
The text was updated successfully, but these errors were encountered: