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Commit54091e6

Browse files
author
zongyanqi
committed
add Easy_202_Happy_Number Easy_263_Ugly_Number Easy_461_Hamming_Distance Easy_476_Number_Complement Easy_520_Detect_Capital.js
1 parent63ae02f commit54091e6

5 files changed

+186
-0
lines changed

‎Easy_202_Happy_Number.js

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/**
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* Write an algorithm to determine if a number is "happy".
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* A happy number is a number defined by the following process: Starting with any positive integer,
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* replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay),
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* or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
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*
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* Example: 19 is a happy number
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*
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* 1^2 + 9^2 = 82
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* 8^2 + 2^2 = 68
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* 6^2 + 8^2 = 100
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* 1^2 + 0^2 + 0^2 = 1
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*
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*/
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/**
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*@param {number} n
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*@return {boolean}
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*/
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varisHappy=function(n){
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varnum=n;
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varresults=[];
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while(true){
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vardigits=(''+num).split('');
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varnewN=squareSumOfDigits(digits);
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if(newN===1)returntrue;
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if(results.indexOf(newN)>-1)returnfalse;
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results.push(newN);
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num=newN;
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}
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};
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functionsquareSumOfDigits(digits){
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returndigits.reduce((sum,d)=>sum+=d*d,0);
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}
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console.log(isHappy(1));
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console.log(isHappy(19));

‎Easy_263_Ugly_Number.js

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/**
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*
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* https://leetcode.com/problems/ugly-number/#/description
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*
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* Write a program to check whether a given number is an ugly number.
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*
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* Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.
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*
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* For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.
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*
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* Note that 1 is typically treated as an ugly number.
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*/
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/**
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*@param {number} num
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*@return {boolean}
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*/
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varisUgly=function(num){
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if(num<=0)returnfalse;
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if(num==1)returntrue;
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while(num>1){
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varold=num;
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if(!(num%2))num=num/2;
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if(!(num%3))num=num/3;
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if(!(num%5))num=num/5;
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if(old===num)returnfalse;
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}
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returntrue;
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};
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console.log(isUgly(6)===true);
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console.log(isUgly(8)===true);
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console.log(isUgly(14)===false);

‎Easy_461_Hamming_Distance.js

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/**
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*
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* The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
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Given two integers x and y, calculate the Hamming distance.
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Note:
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0 ≤ x, y < 231.
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Example:
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Input: x = 1, y = 4
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Output: 2
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Explanation:
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1 (0 0 0 1)
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4 (0 1 0 0)
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↑ ↑
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The above arrows point to positions where the corresponding bits are different.
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*/
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/**
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*@param {number} x
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*@param {number} y
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*@return {number}
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*/
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varhammingDistance=function(x,y){
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varz=x^y;
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varret=0;
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while(z){
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ret+=z%2;
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z=Math.floor(z/2);
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}
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returnret;
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};
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console.log(hammingDistance(1,4));

‎Easy_476_Number_Complement.js

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/**
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*
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*
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* Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
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* Note:
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*
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* The given integer is guaranteed to fit within the range of a 32-bit signed integer.
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* You could assume no leading zero bit in the integer’s binary representation.
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*
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* Example:
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* Input: 5
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* Output: 2
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* Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
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*/
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/**
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*@param {number} num
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*@return {number}
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*/
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varfindComplement=function(num){
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varstr='';
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while(num){
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str=((num%2) ?0 :1)+str;
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num=Math.floor(num/2);
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}
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returnparseInt(str,2);
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};
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console.log(findComplement(5));
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console.log(findComplement(2));

‎Easy_520_Detect_Capital.js

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/**
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* Given a word, you need to judge whether the usage of capitals in it is right or not.
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We define the usage of capitals in a word to be right when one of the following cases holds:
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All letters in this word are capitals, like "USA".
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All letters in this word are not capitals, like "leetcode".
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Only the first letter in this word is capital if it has more than one letter, like "Google".
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Otherwise, we define that this word doesn't use capitals in a right way.
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*/
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/**
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*@param {string} word
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*@return {boolean}
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*/
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vardetectCapitalUse=function(word){
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varlen=word.length;
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varupperLen=0;
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varlowerLen=0;
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varfirstCap=false;
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for(vari=0;i<len;i++){
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varch=word[i];
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if(/[A-Z]/.test(ch)){
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upperLen++;
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if(i==0){
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firstCap=true;
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}
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}
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if(/[a-z]/.test(ch)){
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lowerLen++;
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}
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}
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if(upperLen==len)returntrue;
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if(lowerLen==len)returntrue;
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return(upperLen==1&&firstCap&&len>1);
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};

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