从长度为 M 的无序数组中，找出N个最小的数 All In One

从长度为 M 的无序数组中，找出N个最小的数 All In One

在一组长度为 n 的无序的数组中，取最小的 m个数(m < n), 要求时间复杂度 O(m * n)

网易有道面试题

const minTopK = (m, n) => {
  const obj = {};
  for (let i = 0; i < m.length; i++) {
    if(!obj.hasOwnProperty(m[i])) {
      obj[m[i]] = 1;
    } else {
      obj[m[i]] += 1;
    }
  }
  const arr = Object.entries(obj).sort((a, b) => a[1] - b[1] > 0 ? 1 : -1);
  return arr.slice(0, n).map(item => parseInt(item[0]));
};

demos

如何在 10 亿数中找出前 1000 大的数？

分治法

快速排序 partition

快速排序

O(N*logN)

基本思想：（分治）

1. 先从数列中取出一个数作为key值；

2. 将比这个数小的数全部放在它的左边，大于或等于它的数全部放在它的右边；

3. 对左右两个小数列重复第二步，直至各区间只有1个数。

middle, left, right

const QuickSort = (arr = []) => {
  let m = arr[0];
  let left = [];
  let right = [];
  for(let i of arr) {
   if(i < m ) {
    left.push(i);
   } else {
    right.push(i);
   }
 }
  if(left.length > 1) {
   left = QuickSort(left)
  }
  if(right.length > 1) {
   right = QuickSort(right)
  }
  return left.concat(right);
}

bug

OK

"use strict";

/**
 *
 * @author xgqfrms
 * @license MIT
 * @copyright xgqfrms
 * @created 2020-08-01
 * @modified
 *
 * @description 快速排序 quicksort
 * @difficulty Medium
 * @complexity O(n*log(n))
 * @augments
 * @example
 * @link https://github.com/xgqfrms/leetcode/issues/7#issuecomment-669991209
 * @solutions
 *
 */

const log = console.log;

function quickSort(arr) {
  // 终止条件
  if (arr.length <= 1) {
    return arr;
  }
  // 中间index
  var pivotIndex = Math.floor(arr.length / 2);
  // 中间值，参考值
  var pivot = arr.splice(pivotIndex, 1)[0];
  var left = [];
  var right = [];
  for (var i = 0; i < arr.length; i++) {
    if (arr[i] < pivot) {
      left.push(arr[i]);
    } else {
      right.push(arr[i]);
    }
  }
  // 递归
  return quickSort(left).concat([pivot], quickSort(right));
};

const arr = [12, 7, 5, 23, 18, 37, 1, 9, 17];

const test = quickSort(arr);

log(`arr =\n`, arr)
log(`test =\n`, test)

/*

arr =
 [
  12, 7, 5, 23,
  37, 1, 9, 17
]
test =
 [
   1,  5,  7,  9, 12,
  17, 18, 23, 37
]

*/

leetcode

TopK

https://leetcode.com/problems/top-k-frequent-elements/

https://leetcode.cn/problems/top-k-frequent-elements/

"use strict";

/**
 *
 * @author xgqfrms
 * @license MIT
 * @copyright xgqfrms
 * @created 2020-08-15
 * @modified
 *
 * @description 347. Top K Frequent Elements
 * @difficulty Medium
 * @complexity O(n)
 * @augments
 * @example
 * @link https://leetcode.com/problems/top-k-frequent-elements/
 * @link https://leetcode-cn.com/problems/top-k-frequent-elements/
 * @solutions
 *
 */

const log = console.log;


/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */

var topKFrequent = function(nums, k) {
  const obj = {};
  const result = [];
  for (let i = 0; i < nums.length; i++) {
    if(!obj.hasOwnProperty(nums[i])) {
      obj[nums[i]] = 1;
    } else {
      obj[nums[i]] += 1;
    }
  }
  const arr = Object.entries(obj).sort((a, b) => a[1] - b[1] > 0 ? -1 : 1);
  return arr.slice(0, k).map(item => parseInt(item[0]));
  // return Object.entries(obj).sort((a, b) => a[1] - b[1] > 0 ? -1 : 1).slice(0, k).map(item => parseInt(item[0]));
};

refs

https://www.zhihu.com/question/28874340

https://blog.csdn.net/mashuangwe/article/details/76944143

array sort

https://www.cnblogs.com/xgqfrms/p/12440091.html

facebook hard

https://leetcode.com/problems/alien-dictionary/

refs

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