find-the-odd-int.md 620 B
查找散杂数
题目 ^1
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.
Examples:
[20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5] => 5
[1,1,2,-2,5,2,4,4,-1,-2,5] => -1
思路
数组过滤，单数的特征是首次出现的下标和最后一次出现的下标差值为奇数。
答案：
const findOdd = arr => arr.sort().filter(x=>(arr.lastIndexOf(x)-arr.indexOf(x))%2===0)[0]
其中：
Array.prototype.filter是ES 6的特性。