codewars遇到的一些问题和总结

不定时更新中…

1. Find the odd int

Description:Given an array, find the int that appears an odd number of times.
There will always be only one integer that appears an odd number of times.
这题目乍一看很简单的样子，吾等凡人的想法…循环一下就好…

function findOdd(A) {
  for(var i=0;i

再看看最clever的解法，也是很干净简洁了…

const findOdd = (xs) => xs.reduce((a, b) => a ^ b);

好像恍然大悟的样子…

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关键点就是XOR中:a ^ b = c ， c ^ b = a ，c ^ a = b

eg:
A=[1,2,1,7,8,2,8,2,7]

1^2=3 -------------------------------------->1^2
    3^1=2 ---------------------------------->2
        2^7=5 ------------------------------>2^7
            5^8=13 ------------------------->2^7^8
                13^2=15 -------------------->7^8
                     15^8=7 ---------------->7
                          7^2=5 ------------>7^2
                              5^7=2 -------->2

简单看起来就是：1 ^ 2 ^ 1 ^ 7 ^ 8 ^ 2 ^ 8 ^ 2 ^ 7,出现偶数次的数都抵消掉了，最后剩了个2。这个解法也是由于题目中给定了条件，数组中只有一个整数出现了奇数次。

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2. Roman Numerals Encoder

Description:Create a function taking a positive integer as its parameter and returning a string containing the Roman Numeral representation of that integer.

Modern Roman numerals are written by expressing each digit separately starting with the left most digit and skipping any digit with a value of zero. In Roman numerals 1990 is rendered: 1000=M, 900=CM, 90=XC; resulting in MCMXC. 2008 is written as 2000=MM, 8=VIII; or MMVIII. 1666 uses each Roman symbol in descending order: MDCLXVI.

Example:
solution(1000); // should return ‘M’

Help:
V — 5
X — 10
L — 50
C — 100
D — 500
M — 1,000
Remember that there can’t be more than 3 identical symbols in a row.
More about roman numerals - http://en.wikipedia.org/wiki/Roman_numerals
我一开始的想法是用递归，但是这样做的判断比较多，代码量比较大，感觉没有必要。后来就想用键值对的方式把数据存起来。
solution中clever票数最多的是这个：

function solution(number){
  // convert the number to a roman numeral
var  roman = {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 }

var ans = '';
while(number>0){
    for(a in roman){ 
        if(roman[a]<=number){ ans += a; number-=roman[a]; break;}
            
    }
}
return ans;
}

这个题依赖于键值对的顺序，在遍历的时候，要依照1000,900,500…这样的顺序遍历。但是对象是无序的，for-in 无法保证遍历顺序。所以我觉得这里用数组存数据会好一点。

function solution(number){
    // convert the number to a roman numeral
  decimals = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
  roman    = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
  var ans = '';
  while(number>0){
      for(index in decimals){ 
          if(decimals[index]<=number){ ans += roman[index]; number-=decimals[index]; break;}
      }
  }
  return ans;
}