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top-k-frequent-elements solution
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/**
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* @param {number[]} nums
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* @param {number} k
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* @return {number[]}
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*/
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var topKFrequent = function(nums, k) {
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let dict = new Map();
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let temp = Array.from(new Set([...nums]));
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for(const num of nums){
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if(dict.has(num)){
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dict.set(num, dict.get(num)+1);
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}else{
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dict.set(num,1);
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}
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}
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temp = temp.sort((a,b)=>{
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let aCount = dict.get(a);
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let bCount = dict.get(b);
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return bCount - aCount;
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})
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let slice = temp.slice(0,k);
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return slice;
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};

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