js 二进制 & 位运算 All In One

js 二进制 & 位运算 All In One

binary & bitwise

位异或 ^

XOR

n^0 === n；任何数字 n 与 0 异或，等于自己；

n^n === 0；任何数字 n 与自己异或，等于 0

a^bc === a^cb；异或运算具有交换律

// 异或 === 先异后或 ✅

// 先或后异

3 ^ 3;
// 0

3 ^ 0;
// 3

3 === 0b00000011;
// true
0 === 0b00000000;
// true

异或: 运算的两个数子转换成二进制后，对应位上的两位数只有一个是 1 才会返回 1；全 0 或 全 1 都会返回 0; ✅

- 3 === 0b00000011;
- 0 === 0b00000000;
// 异或 ✅
+ 3 === 0b00000011;


// 1 + 0 => 1 ✅
// 1 + 1 => 0
// 0 + 0 => 0

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators#binary_bitwise_operators

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_XOR

const a = 5;        // 00000000000000000000000000000101
const b = 3;        // 00000000000000000000000000000011
console.log(a ^ b); // 00000000000000000000000000000110
// 6
const x = 0;        // 000000000
const y = 3;        // 000000011
console.log(x ^ y); // 000000011
// 3
const u = 1;        // 000000001
const f = 3;        // 000000011
console.log(u ^ f); // 000000010
// 2

Bit Manipulation Concept

If we take XOR of zero and some bit, it will return that bit
a \oplus 0 = aa⊕0=a
If we take XOR of two same bits, it will return 0
a \oplus a = 0a⊕a=0
a \oplus b \oplus a = (a \oplus a) \oplus b = 0 \oplus b = ba⊕b⊕a=(a⊕a)⊕b=0⊕b=b
So we can XOR all bits together to find the unique number.

"use strict";

/**
 *
 * @author xgqfrms
 * @license MIT
 * @copyright xgqfrms
 * @created 2020-08-15
 * @modified
 *
 * @description 136 single-number
 * @difficulty Easy
 * @complexity O(n)
 * @augments
 * @example
 * @link https://leetcode.cn/problems/single-number/
 * @link https://leetcode.cn/leetbook/read/top-interview-questions/xm0u83/
 * @solutions
 *
 */

const log = console.log;


/**
 * @param {number[]} nums
 * @return {number}
 */
var singleNumber = function(nums) {
  let a = 0;
  for (const i of nums) {
    // 异或运算
    a ^= i;
  }
  return a;
};
// Time complexity : O(n)
// Space complexity : O(1)


// var singleNumber = function(nums) {
//   return nums.reduce((sum, i) => sum ^ i, 0);
// };

/*

输入: [2,2,1]
输出: 1

输入: [4,1,2,1,2]
输出: 4

*/


const test = [2,2,1];
const result = singleNumber(test);
log(`result =`, result)

const test2 = [4,1,2,1,2];
const result2 = singleNumber(test2);
log(`result2 =`, result2)

移位运算

"use strict";

/**
 *
 * @author xgqfrms
 * @license MIT
 * @copyright xgqfrms
 * @created 2019-08-14
 *
 * @description auto-sevent-times-without-math.js
 * @description 不用加减乘除运算符, 求整数的7倍
 * @description js array get sum value without call math methods
 * @augments
 * @example eval([1,2,3].join('+'));// 6
 * @link https://stackoverflow.com/questions/1230233/how-to-find-the-sum-of-an-array-of-numbers
 *
 */

let log = console.log;

const autoSevenTimes = (int = 0, times = 7, debug = false) => {
    let result = [];
    if (!int) {
        return 0;
    } else {
        for (let i = 0; i < times; i++) {
            result.push(int);
        }
    }
    // result = result.reduce((acc, item) => acc + item);
    result = eval(result.join(`+`));
    if(debug) {
        log(`result = `, result);
    }
    return result;
};

let num = 3;
autoSevenTimes(num);
// expect: 3 * 7 = 21

const autoSeventTimes = (num = 0, times = 7) => {
    let x = Math.floor(times / 2);
    // 左移 n 位操作，倍增 2^n 倍
    // 左移 3 位，扩大8 倍(2 ** 3)；倍增后，减去一次自身， 等于 7 倍
    return (num << x) - num;
};

let xyz = autoSeventTimes(3);
// 21

console.log(`xyz =`, xyz);

// 3 === 00000011
// 24 === 00011000

// 21 === 00010101


// 0b 二进制
// 0o 十六进制
let x = 0b00010101;
// 21

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Numbers_and_dates#binary_numbers

位运算 / Bitwise Operators

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Expressions_and_Operators

refs

https://github.com/xgqfrms/learning-javascript-with-mdn/issues/8

https://repl.it/@xgqfrms/bitwise-operator-and-left-shift-and-right-shift

js 大数相加计算解析 All In One

https://www.cnblogs.com/xgqfrms/p/12995736.html

js & bitwise operator

https://www.cnblogs.com/xgqfrms/p/11563020.html

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