题目： 找出数组中出现奇数次数的数字

描述： Given an array of integers, find the one that appears an odd number of times. There will always be only one integer that appears an odd number of times. Examples [7] should return 7, because it occurs 1 time (which is odd). [0] should return 0, because it occurs 1 time (which is odd). [1,1,2] should return 2, because it occurs 1 time (which is odd). [0,1,0,1,0] should return 0, because it occurs 3 times (which is odd). [1,2,2,3,3,3,4,3,3,3,2,2,1] should return 4, because it appears 1 time (which is odd).

在线地址：Find the odd int

基础实现：统计每个数字出现的次数找出奇数次数的数字即可

function findOdd(A) {
  var obj = {};
  A.forEach(function(el){
    obj[el] ? obj[el]++ : obj[el] = 1;
  });
  
  for(prop in obj) {
    if(obj[prop] % 2 !== 0) return Number(prop);
  }
}

看到这个问题，一般人想到的都是上面的解决方案，下面欣赏下网站上大佬给出的最佳实践： 利用异或运算符^的原理：相同数字异或结果为0，0与任意数异或为任意数；把所有数字依次异或即可剔出偶数次数的数字

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