347. Top K Frequent Elements 出现频率最高的k个元素

 Given a non-empty array of integers, return the k most frequent elements.

For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2].

Note:

• You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
• Your algorithm's time complexity must be better than O(n log n), where n is the array's size.

例如，给定[1,1,1,2,2,3]和k = 2，返回[1,2]。
注意：
你可以假设k总是有效的，1≤k≤唯一元素的个数。
你的算法的时间复杂度必须比O（n log n）好，其中n是数组的大小。

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
let left = (i) => 2 * i + 1;
let right = (i) => 2 * (i + 1);
let buildHeap = (x, less_p) => {
    let n = x.length;
    for (let i = Math.floor(n / 2) - 1; i >= 0; i--) {
        heapify(x, i, less_p)
    }
}
let heapify = (x, i, less_p) => {
    let n = x.length;
    while (true) {
        let l = left(i);
        let r = right(i);
        let smallest = i;
        if (l < n && less_p(x[l], x[i])) {
            smallest = l;
        }
        if (r < n && less_p(x[r], x[smallest])) {
            smallest = r;
        }
        if (smallest != i) {
            [x[i], x[smallest]] = [x[smallest], x[i]];
            i = smallest;
        } else {
            break;
        }
    }
}
let heapPop = (x, less_p) => {
    top = x[0]
    x[0] = x[x.length - 1];
    x.pop()
    if (x.length > 0) {
        heapify(x, 0, less_p)
    }
    return top
}
let topK = (x, k, less_p) => {
    let res = [];
    buildHeap(x, less_p);
    let count = x.length;
    for (let i = 0; i < Math.min(k, count); i++) {
        res.push(heapPop(x, less_p));
    }
    return res;
}
var topKFrequent = function (nums, k) {
    let MAX_HEAP = (a, b) => m[a] > m[b];
    let m = {};
    let heap = [];
    for (let i in nums) {
        let c = nums[i];
        !m[c] && heap.push(c);
        m[c] = m[c] ? ++m[c] : 1;
    }
    return topK(heap, k, MAX_HEAP);
};
let nums = [1, 2];
let k = 2;
let res = topKFrequent(nums, k);
console.log(res);

来自为知笔记(Wiz)