code wars 笔记2

caketi

[JavaScript] 纯文本查看 复制代码

function solution(number){
  let numerals = {
    "M": 1000,
    "CM": 900,
    "D": 500,
    "CD": 400,
    "C": 100,
    "XC": 90,
    "L": 50,
    "XL": 40,
    "X": 10,
    "IX": 9,
    "V": 5,
    "IV": 4,
    "I": 1
  }, result = '';
  
  Object.keys(numerals).forEach(e => {
    let n = Math.floor(number / numerals[e]);
    number -= n * numerals[e];
    result += e.repeat(n);
  });
  
  return result;    
}

Object. keys() returns an array whose elements are strings corresponding to the enumerable properties found directly upon object . The ordering of the properties is the same as that given by looping over the properties of the object manually. --- 返回的数组---有点map的味道--一一对应
The Math. floor() function returns the largest integer less than or equal to a given number.---返回小于等于


function romanize (num) {    if (isNaN(num))        return NaN;    var digits = String(+num).split(""),        key = ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",               "","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",               "","I","II","III","IV","V","VI","VII","VIII","IX"],        roman = "",        i = 3;    while (i--)        roman = (key[+digits.pop() + (i * 10)] || "") + roman;    return Array(+digits.join("") + 1).join("M") + roman;}

[JavaScript] 纯文本查看 复制代码

function romanize (num) {    if (isNaN(num))
        return NaN;
    var digits = String(+num).split(""),
        key = ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
               "","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
               "","I","II","III","IV","V","VI","VII","VIII","IX"],
        roman = "",
        i = 3;
    while (i--)
        roman = (key[+digits.pop() + (i * 10)] || "") + roman;
    return Array(+digits.join("") + 1).join("M") + roman;
}

[JavaScript] 纯文本查看 复制代码

var numeralCodes = [["","I","II","III","IV","V","VI","VII","VIII","IX"],         // Ones
                    ["","X","XX","XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"],   // Tens
                    ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"]];        // Hundreds

function convert(num) {
  var numeral = "";
  var digits = num.toString().split('').reverse();
  for (var i=0; i < digits.length; i++){
    numeral = numeralCodes[i][parseInt(digits[i])] + numeral;
  }
  return numeral;  
}

[Asm] 纯文本查看 复制代码

function convertToRoman(num) {  //create key:value pairs
  var romanLookup = {M:1000, D:500, C:100, L:50, X:10, V:5, I:1};
  var roman = [];
  var romanKeys = Object.keys(romanLookup);
  var curValue;
  var index;
  var count = 1;

  for(var numeral in romanLookup){
    curValue = romanLookup[numeral];
    index = romanKeys.indexOf(numeral);

    while(num >= curValue){

      if(count < 4){
        //push up to 3 of the same numeral
        roman.push(numeral);
      } else {
        //else we had to push four, so we need to convert the numerals 
        //to the next highest denomination "coloring-up in poker speak"

        //Note: We need to check previous index because it might be part of the current number.
        //Example:(9) would attempt (VIIII) so we would need to remove the V as well as the I's
        //otherwise removing just the last three III would be incorrect, because the swap 
        //would give us (VIX) instead of the correct answer (IX)
        if(roman.indexOf(romanKeys[index - 1]) > -1){
          //remove the previous numeral we worked with 
          //and everything after it since we will replace them
          roman.splice(roman.indexOf(romanKeys[index - 1]));
          //push the current numeral and the one that appeared two iterations ago; 
          //think (IX) where we skip (V)
          roman.push(romanKeys[index], romanKeys[index - 2]);
        } else {
          //else Example:(4) would attemt (IIII) so remove three I's and replace with a V 
          //to get the correct answer of (IV)

          //remove the last 3 numerals which are all the same
          roman.splice(-3);
          //push the current numeral and the one that appeared right before it; think (IV)
          roman.push(romanKeys[index], romanKeys[index - 1]);
        }
      }
      //reduce our number by the value we already converted to a numeral
      num -= curValue;
      count++;
    }
    count = 1;
  }
  return roman.join("");
}

convertToRoman(36);