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).
function findOdd(A) {
let newObj = A.reduce((obj, el) => {
if (el in obj) obj[el] += 1;
else obj[el] = 1;
return obj;
}, {})
for (let el in newObj) {
if (newObj[el] % 2 === 1)
return Number(el);
}
return 0;
}
같은 풀이인데도 이렇게 짧게 줄여쓸 수도 있다.
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);
}
}