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October 22, 2017 03:40
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/*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. | |
Note: | |
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. | |
*/ | |
var topKFrequent = function(words, k) { | |
var map = {}; | |
words.forEach(function(item){ | |
if(map[item]){ | |
map[item] = map[item]+1; | |
} else { | |
map[item] = 1; | |
} | |
}); | |
var arr = []; | |
for(key in map) { | |
arr.push({key: key, val: map[key]}); | |
} | |
return arr.sort(function(a, b){ | |
if(a.val===b.val) { | |
return a.key.localeCompare(b.key); | |
} | |
return b.val-a.val; | |
}).slice(0,k).map(function(item){ | |
return item.key | |
}); | |
}; |
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