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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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var dict = {};
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var sortTable = [];
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for(let i = 0; i < nums.length; i ++) {
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dict[nums[i]] = dict[nums[i]] ? dict[nums[i]] + 1 : 1;
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}
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for(let ele in dict) {
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sortTable.push([ele, dict[ele]]);
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}
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sortTable.sort((a, b) => {
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return b[1] - a[1];
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});
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return sortTable.slice(0, k).map(e => Number(e[0]))
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};

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