JavaScript 面試中常見算法問題詳解

原帖地址：http://blog.csdn.net/wky_csdn/article/details/66973737

JavaScript 面試中常見算法問題詳解翻譯自 Interview Algorithm Questions in Javascript 從屬於筆者的 Web 前端入門與工程實踐。下文提到的很多問題從算法角度並不一定多麼困難，不過用 JavaScript 內置的 API 來完成還是需要一番考量的。

JavaScript Specification

闡述下 JavaScript 中的變量提升

所謂提升，顧名思義即是 JavaScript 會將所有的聲明提升到當前作用域的頂部。這也就意味着我們可以在某個變量聲明前就使用該變量，不過雖然 JavaScript 會將聲明提升到頂部，但是並不會執行真的初始化過程。

闡述下 use strict; 的作用

use strict; 顧名思義也就是 JavaScript 會在所謂嚴格模式下執行，其一個主要的優勢在於能夠強制開發者避免使用未聲明的變量。對於老版本的瀏覽器或者執行引擎則會自動忽略該指令。

// Example of strict mode
“use strict”;

catchThemAll();
function catchThemAll() {
x = 3.14; // Error will be thrown
return x * x;
}

解釋下什麼是 Event Bubbling 以及如何避免

Event Bubbling 即指某個事件不僅會觸發當前元素，還會以嵌套順序傳遞到父元素中。直觀而言就是對於某個子元素的點擊事件同樣會被父元素的點擊事件處理器捕獲。避免 Event Bubbling 的方式可以使用event.stopPropagation() 或者 IE 9 以下使用event.cancelBubble。

== 與 === 的區別是什麼

=== 也就是所謂的嚴格比較，關鍵的區別在於=== 會同時比較類型與值，而不是僅比較值。

// Example of comparators
0 == false; // true
0 === false; // false

2 == ‘2’; // true
2 === ‘2’; // false

解釋下 null 與 undefined 的區別

JavaScript 中，null 是一個可以被分配的值，設置爲 null 的變量意味着其無值。而 undefined 則代表着某個變量雖然聲明瞭但是尚未進行過任何賦值。

解釋下 Prototypal Inheritance 與 Classical Inheritance 的區別

在類繼承中，類是不可變的，不同的語言中對於多繼承的支持也不一樣，有些語言中還支持接口、final、abstract 的概念。而原型繼承則更爲靈活，原型本身是可以可變的，並且對象可能繼承自多個原型。

數組

找出整型數組中乘積最大的三個數

給定一個包含整數的無序數組，要求找出乘積最大的三個數。

var unsorted_array = [-10, 7, 29, 30, 5, -10, -70];

computeProduct(unsorted_array); // 21000

function sortIntegers(a, b) {
return a - b;
}

// greatest product is either (min1 * min2 * max1 || max1 * max2 * max3)
function computeProduct(unsorted) {
var sorted_array = unsorted.sort(sortIntegers),
product1 = 1,
product2 = 1,
array_n_element = sorted_array.length - 1;

// Get the product of three largest integers in sorted array
for (var x = array_n_element; x > array_n_element - 3; x–) {
product1 = product1 * sorted_array[x];
}
product2 = sorted_array[0] * sorted_array[1] * sorted_array[array_n_element];

if (product1 > product2) return product1;

return product2
};

尋找連續數組中的缺失數

給定某無序數組，其包含了 n 個連續數字中的 n – 1 個，已知上下邊界，要求以O(n)的複雜度找出缺失的數字。

// The output of the function should be 8
var array_of_integers = [2, 5, 1, 4, 9, 6, 3, 7];
var upper_bound = 9;
var lower_bound = 1;

findMissingNumber(array_of_integers, upper_bound, lower_bound); //8

function findMissingNumber(array_of_integers, upper_bound, lower_bound) {

// Iterate through array to find the sum of the numbers
var sum_of_integers = 0;
for (var i = 0; i < array_of_integers.length; i++) {
sum_of_integers += array_of_integers[i];
}

// 以高斯求和公式計算理論上的數組和
// Formula: [(N * (N + 1)) / 2] - [(M * (M - 1)) / 2];
// N is the upper bound and M is the lower bound

upper_limit_sum = (upper_bound * (upper_bound + 1)) / 2;
lower_limit_sum = (lower_bound * (lower_bound - 1)) / 2;

theoretical_sum = upper_limit_sum - lower_limit_sum;

//
return (theoretical_sum - sum_of_integers)
}

數組去重

給定某無序數組，要求去除數組中的重複數字並且返回新的無重複數組。

// ES6 Implementation
var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];

Array.from(new Set(array)); // [1, 2, 3, 5, 9, 8]

// ES5 Implementation
var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];

uniqueArray(array); // [1, 2, 3, 5, 9, 8]

數組中元素最大差值計算

給定某無序數組，求取任意兩個元素之間的最大差值，注意，這裏要求差值計算中較小的元素下標必須小於較大元素的下標。譬如[7, 8, 4, 9, 9, 15, 3, 1, 10]這個數組的計算值是 11( 15 – 4 ) 而不是 14(15 – 1)，因爲 15 的下標小於 1。

var array = [7, 8, 4, 9, 9, 15, 3, 1, 10];
// [7, 8, 4, 9, 9, 15, 3, 1, 10] would return 11 based on the difference between 4 and 15
// Notice: It is not 14 from the difference between 15 and 1 because 15 comes before 1.

findLargestDifference(array);

function findLargestDifference(array) {

// 如果數組僅有一個元素，則直接返回 -1

if (array.length <= 1) return -1;

// current_min 指向當前的最小值

var current_min = array[0];
var current_max_difference = 0;

// 遍歷整個數組以求取當前最大差值，如果發現某個最大差值，則將新的值覆蓋 current_max_difference
// 同時也會追蹤當前數組中的最小值，從而保證 largest value in future - smallest value before it

for (var i = 1; i < array.length; i++) {
if (array[i] > current_min && (array[i] - current_min > current_max_difference)) {
current_max_difference = array[i] - current_min;
} else if (array[i] <= current_min) {
current_min = array[i];
}
}

// If negative or 0, there is no largest difference
if (current_max_difference <= 0) return -1;

return current_max_difference;
}

數組中元素乘積

給定某無序數組，要求返回新數組 output ，其中 output[i] 爲原數組中除了下標爲 i 的元素之外的元素乘積，要求以 O(n) 複雜度實現：

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

productExceptSelf(firstArray); // [8, 8, 4, 16]
productExceptSelf(secondArray); // [0, 0, 0, 0]
productExceptSelf(thirdArray); // [12, 12, 8, -12]

function productExceptSelf(numArray) {
var product = 1;
var size = numArray.length;
var output = [];

// From first array: [1, 2, 4, 16]
// The last number in this case is already in the right spot (allows for us)
// to just multiply by 1 in the next step.
// This step essentially gets the product to the left of the index at index + 1
for (var x = 0; x < size; x++) {
output.push(product);
product = product * numArray[x];
}

// From the back, we multiply the current output element (which represents the product
// on the left of the index, and multiplies it by the product on the right of the element)
var product = 1;
for (var i = size - 1; i > -1; i–) {
output[i] = output[i] * product;
product = product * numArray[i];
}

return output;
}

數組交集

給定兩個數組，要求求出兩個數組的交集，注意，交集中的元素應該是唯一的。

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

intersection(firstArray, secondArray); // [2, 1]

function intersection(firstArray, secondArray) {
// The logic here is to create a hashmap with the elements of the firstArray as the keys.
// After that, you can use the hashmap’s O(1) look up time to check if the element exists in the hash
// If it does exist, add that element to the new array.

var hashmap = {};
var intersectionArray = [];

firstArray.forEach(function(element) {
hashmap[element] = 1;
});

// Since we only want to push unique elements in our case… we can implement a counter to keep track of what we already added
secondArray.forEach(function(element) {
if (hashmap[element] === 1) {
intersectionArray.push(element);
hashmap[element]++;
}
});

return intersectionArray;

// Time complexity O(n), Space complexity O(n)
}

字符串

顛倒字符串

給定某個字符串，要求將其中單詞倒轉之後然後輸出，譬如”Welcome to this Javascript Guide!” 應該輸出爲 “emocleW ot siht tpircsavaJ !ediuG”。

var string = “Welcome to this Javascript Guide!”;

// Output becomes !ediuG tpircsavaJ siht ot emocleW
var reverseEntireSentence = reverseBySeparator(string, “”);

// Output becomes emocleW ot siht tpircsavaJ !ediuG
var reverseEachWord = reverseBySeparator(reverseEntireSentence, ” “);

function reverseBySeparator(string, separator) {
return string.split(separator).reverse().join(separator);
}

亂序同字母字符串

給定兩個字符串，判斷是否顛倒字母而成的字符串，譬如Mary與Army就是同字母而順序顛倒：

var firstWord = “Mary”;
var secondWord = “Army”;

isAnagram(firstWord, secondWord); // true

function isAnagram(first, second) {
// For case insensitivity, change both words to lowercase.
var a = first.toLowerCase();
var b = second.toLowerCase();

// Sort the strings, and join the resulting array to a string. Compare the results
a = a.split(“”).sort().join(“”);
b = b.split(“”).sort().join(“”);

return a === b;
}

會問字符串

判斷某個字符串是否爲迴文字符串，譬如racecar與race car都是迴文字符串：

isPalindrome(“racecar”); // true
isPalindrome(“race Car”); // true

function isPalindrome(word) {
// Replace all non-letter chars with “” and change to lowercase
var lettersOnly = word.toLowerCase().replace(/\s/g, “”);

// Compare the string with the reversed version of the string
return lettersOnly === lettersOnly.split(“”).reverse().join(“”);
}

棧與隊列

使用兩個棧實現入隊與出隊

var inputStack = []; // First stack
var outputStack = []; // Second stack

// For enqueue, just push the item into the first stack
function enqueue(stackInput, item) {
return stackInput.push(item);
}

function dequeue(stackInput, stackOutput) {
// Reverse the stack such that the first element of the output stack is the
// last element of the input stack. After that, pop the top of the output to
// get the first element that was ever pushed into the input stack
if (stackOutput.length <= 0) {
while(stackInput.length > 0) {
var elementToOutput = stackInput.pop();
stackOutput.push(elementToOutput);
}
}

return stackOutput.pop();
}

判斷大括號是否閉合

創建一個函數來判斷給定的表達式中的大括號是否閉合：

var expression = “{{}}{}{}”
var expressionFalse = “{}{{}”;

isBalanced(expression); // true
isBalanced(expressionFalse); // false
isBalanced(“”); // true

function isBalanced(expression) {
var checkString = expression;
var stack = [];

// If empty, parentheses are technically balanced
if (checkString.length <= 0) return true;

for (var i = 0; i < checkString.length; i++) {
if(checkString[i] === ‘{‘) {
stack.push(checkString[i]);
} else if (checkString[i] === ‘}’) {
// Pop on an empty array is undefined
if (stack.length > 0) {
stack.pop();
} else {
return false;
}
}
}

// If the array is not empty, it is not balanced
if (stack.pop()) return false;
return true;
}

遞歸

二進制轉換

通過某個遞歸函數將輸入的數字轉化爲二進制字符串：

decimalToBinary(3); // 11
decimalToBinary(8); // 1000
decimalToBinary(1000); // 1111101000

二分搜索

function recursiveBinarySearch(array, value, leftPosition, rightPosition) {
// Value DNE
if (leftPosition > rightPosition) return -1;

數字

判斷是否爲 2 的指數值

isPowerOfTwo(4); // true
isPowerOfTwo(64); // true
isPowerOfTwo(1); // true
isPowerOfTwo(0); // false
isPowerOfTwo(-1); // false

// For the non-zero case:
function isPowerOfTwo(number) {
// & uses the bitwise n.
// In the case of number = 4; the expression would be identical to:
// return (4 & 3 === 0)
// In bitwise, 4 is 100, and 3 is 011. Using &, if two values at the same
// spot is 1, then result is 1, else 0. In this case, it would return 000,
// and thus, 4 satisfies are expression.
// In turn, if the expression is return (5 & 4 === 0), it would be false
// since it returns 101 & 100 = 100 (NOT === 0)

return number & (number - 1) === 0;
}

// For zero-case:
function isPowerOfTwoZeroCase(number) {
return (number !== 0) && ((number & (number - 1)) === 0);
}

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