每日JavaScript - 3

最新推荐文章于 2025-04-26 23:57:28 发布

原创最新推荐文章于 2025-04-26 23:57:28 发布 · 181 阅读

0 ·

CC 4.0 BY-SA版权

前端同时被 2 个专栏收录

62 篇文章

订阅专栏

Javascript

44 篇文章

订阅专栏

本文探讨了在O(n)复杂度下解决特定算法问题的方法，包括寻找缺失的自然数、计算数组中最大差值、生成除自身外的乘积数组以及查找数组交集。通过具体实例和代码实现，展示了如何运用数学原理和编程技巧优化算法效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

给定一个乱序而且缺少一个值的连续自然数数组如下面代码的arrayOfIntegers，求再O(n)下找到缺少的值。
说来惭愧，初步思考有点摸不清思路，看了代码后才明白这不就是初中数学题码？
原答案：

// The output of the function should be 8
var arrayOfIntegers = [2, 5, 1, 4, 9, 6, 3, 7];
var upperBound = 9;
var lowerBound = 1;

console.log(findMissingNumber(arrayOfIntegers, upperBound, lowerBound)); // 8

function findMissingNumber(arrayOfIntegers, upperBound, lowerBound) {
  // Iterate through array to find the sum of the numbers
  var sumOfIntegers = 0;
  for (var i = 0; i < arrayOfIntegers.length; i++) {
    sumOfIntegers += arrayOfIntegers[i];
  }

  // Find theoretical sum of the consecutive numbers using a variation of Gauss Sum.
  // Formula: [(N * (N + 1)) / 2] - [(M * (M - 1)) / 2];
  // N is the upper bound and M is the lower bound

  upperLimitSum = (upperBound * (upperBound + 1)) / 2;
  lowerLimitSum = (lowerBound * (lowerBound - 1)) / 2;

  theoreticalSum = upperLimitSum - lowerLimitSum;

  return theoreticalSum - sumOfIntegers;
}

但是给了首项和末项，也可以很快用高斯求和。

function missing (arr, up, low) {
	// 项数
	let long = up - low + 1;
	let gaussSum = (low + up) * long / 2;
	let len = arr.length;
	let sum = 0;
	for (let i = 0; i < arr.length; i++) {
		sum += arr[i];
	}
	return gaussSum - sum;
}

console.log(missing(arrayOfIntegers, upperBound , lowerBound));

给定一个数组，求后面元素减去前面元素的最大差值

var array = [7, 8, 4, 9, 9, 15, 3, 1, 10];

function max(arr) {
	let len =  arr.length;
	let newArr = [];
	for (let i = 0; i < len; i++) {
		let temp = arr[i];
		for (let e = i + 1; e < len; e++) {
			if (arr[e] > temp) {
				temp = arr[e]
			}
            newArr.push(temp - arr[i]);
		}
	}
	newArr.sort((a,b) => b - a);
	return newArr[0];
}

console.log(max(array))

可以不用newArr，存结果，直接更新最大差值就可以，我觉得这个更符合我的思维就这样写了。
Given an array of integers, return an output array such that output[i] is equal to the product of all the elements in the array other than itself. (Solve this in O(n) without division)

var firstArray = [2, 2, 4, 1];
var secondArray = [0, 0, 0, 2];
var thirdArray = [-2, -2, -3, 2];

function except(arr) {
	let res = [];
	let temp = 1;
	let len = arr.length;
	for (let i = 0; i < len; i++) {
		for (let e = 0; e < len; e++) {
			if (e != i) {
				temp = temp * arr[e];
			}
		}
		res.push(temp);
		// 一个循环结束把变量变为1
		temp = 1;
	}
	return res;
}
console.log(except(firstArray));
console.log(except(secondArray));
console.log(except(thirdArray));

找数组的交集，非常简单，但太晚了，我就随便谢谢。

var firstArray = [2, 2, 4, 1];
var secondArray = [1, 2, 0, 2];

function intersection(arr1, arr2) {
	let res = [];
	let len = arr1.length;
	for (let i = 0; i < len; i++) {
		if(arr2.includes(arr1[i])) {
			res.push(arr1[i]);
		}
	}
	res = [...new Set(res)];
	return res;
}

console.log(intersection(firstArray, secondArray));