我正在尝试解决这个练习，找出在数组中出现奇数次数的练习。到目前为止我有这个，但输出最终是一个出现偶数次的整数。例如，数字2出现3次，数字4出现6次，但输出为4，因为它将其计为出现5次。怎么能将它发现的第一个集合返回为奇数？任何帮助表示赞赏！
function oddInt(array) {
var count = 0;
var element = 0;
for(var i = 0; i < array.length; i++) {
var tempInt = array[i];
var tempCount = 0;
for(var j = 0; j <array.length; j++) {
if(array[j]===tempInt) {
tempCount++;
if(tempCount % 2 !== 0 && tempCount > count) {
count = tempCount;
element = array[j];
}
}
}
}
return element;
}
oddInt([1,2,2,2,4,4,4,4,4,4,5,5]);
答案 0 :(得分：4)
function findOdd(numbers) {
var count = 0;
for(var i = 0; i<numbers.length; i++){
for(var j = 0; j<numbers.length; j++){
if(numbers[i] == numbers[j]){
count++;
}
}
if(count % 2 != 0 ){
return numbers[i];
}
}
};
console.log(findOdd([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5])); //5
console.log(findOdd([1,1,1,1,1,1,10,1,1,1,1])); //10
答案 1 :(得分：2)
function oddInt(array) {
// first: let's count occurences of all the elements in the array
var hash = {}; // object to serve as counter for all the items in the array (the items will be the keys, the counts will be the values)
array.forEach(function(e) { // for each item e in the array
if(hash[e]) hash[e]++; // if we already encountered this item, then increments the counter
else hash[e] = 1; // otherwise start a new counter (initialized with 1)
});
// second: we select only the numbers that occured an odd number of times
var result = []; // the result array
for(var e in hash) { // for each key e in the hash (the key are the items of the array)
if(hash[e] % 2) // if the count of that item is an odd number
result.push(+e); // then push the item into the result array (since they are keys are strings we have to cast them into numbers using unary +)
}
return result;
}
console.log(oddInt([1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 5, 5]));
仅返回第一个：
function oddInt(array) {
var hash = {};
array.forEach(function(e) {
if(hash[e]) hash[e]++;
else hash[e] = 1;
});
for(var e in hash) { // for each item e in the hash
if(hash[e] % 2) // if this number occured an odd number of times
return +e; // return it and stop looking for others
}
// default return value here
}
console.log(oddInt([1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 5, 5]));
答案 2 :(得分：2)
首先找到频率，然后找出哪些是奇数：
const data = [1,2,2,2,4,4,4,4,4,4,5,5]
const freq = data.reduce(
(o, k) => ({ ...o, [k]: (o[k] || 0) + 1 }),
{})
const oddFreq = Object.keys(freq).filter(k => freq[k] % 2)
// => ["1", "2"]
答案 3 :(得分：0)
这是因为当检查第5个if(tempCount % 2 !== 0 && tempCount > count)时，您的条件4为真。这会更新count和element变量。
选中第6个4时，条件为假。
要修复，请将条件移到最里面的循环之外，这样只有在计算了数组中的所有数字后才能检查它。
答案 4 :(得分：0)
function oddInt(array, minCount, returnOne) {
minCount = minCount || 1;
var itemCount = array.reduce(function(a, b) {
a[b] = (a[b] || 0) + 1;
return a;
}, {});
/*
itemCount: {
"1": 1,
"2": 3,
"4": 6,
"5": 2,
"7": 3
}
*/
var values = Object.keys(itemCount).filter(function(k) {
return itemCount[k] % 2 !== 0 && itemCount[k]>=minCount;
});
return returnOne?values[0]:values;
}
var input = [1, 2, 2, 2, 4, 4, 4, 4, 4, 4, 5, 5, 7, 7, 7];
console.log(oddInt(input, 3, true));
console.log(oddInt(input, 1, true));
console.log(oddInt(input, 2, false));
答案 5 :(得分：0)
之所以发生这种情况，是因为每次找到奇数时都会设置element变量，所以当它找到一个，三个和五个4时就会设置它。
让我们一步一步检查代码：
function oddInt(array) {
// Set the variables. The count and the element, that is going to be the output
var count = 0;
var element = 0;
// Start looking the array
for(var i = 0; i < array.length; i++) {
// Get the number to look for and restart the tempCount variable
var tempInt = array[i];
var tempCount = 0;
console.log("");
console.log(" * Looking for number", tempInt);
// Start looking the array again for the number to look for
for(var j = 0; j <array.length; j++) {
// If the current number is the same as the one that we are looking for, sum it up
console.log("Current number at position", j, "is", array[j]);
if(array[j]===tempInt) {
tempCount++;
console.log("Number found. Current count is", tempCount);
// Then, if currently there are an odd number of elements, save the number
// Note that you are calling this altough you don't have looped throgh all the array, so the console will log 3 and 5 for the number '4'
if(tempCount % 2 !== 0 && tempCount > count) {
console.log("Odd count found:", tempCount);
count = tempCount;
element = array[j];
}
}
}
}
return element;
}
oddInt([1,2,2,2,4,4,4,4,4,4,5,5]);
我们要做的是检查循环所有数组后的计数，这样：
function oddInt(array) {
// Set the variables. The count and the element, that is going to be the output
var count = 0;
var element = 0;
// Start looking the array
for(var i = 0; i < array.length; i++) {
// Get the number to look for and restart the tempCount variable
var tempInt = array[i];
var tempCount = 0;
console.log("");
console.log(" * Looking for number", tempInt);
// Start looking the array again for the number to look for
for(var j = 0; j <array.length; j++) {
// If the current number is the same as the one that we are looking for, sum it up
console.log("Current number at position", j, "is", array[j]);
if(array[j]===tempInt) {
tempCount++;
console.log("Number found. Current count is", tempCount);
}
}
// After getting all the numbers, then we check the count
if(tempCount % 2 !== 0 && tempCount > count) {
console.log("Odd count found:", tempCount);
count = tempCount;
element = tempInt;
}
}
return element;
}
oddInt([1,2,2,2,4,4,4,4,4,4,5,5]);
顺便说一下，这只是为了让你了解问题的位置并从中吸取教训，虽然这不是最优化的方法，因为你可能会注意到你正在寻找，比方说，数字{ {1}}三次，当你第一次得到你想要的输出时。如果表演是家庭作业的一部分，那么你应该考虑另一种方式：P
答案 6 :(得分：0)
“A”是要检查的数组。
function findOdd(A) {
var num;
var count =0;
for(i=0;i<A.length;i++){
num = A[i]
for(a=0;a,a<A.length;a++){
if(A[a]==num){
count++;
}
} if(count%2!=0){
return num;
}
}
}
答案 7 :(得分：0)
function oddOne (sorted) {
let temp = sorted[0];
let count = 0;
for (var i = 0; i < sorted.length; i++) {
if (temp === sorted[i]) {
count++;
if (i === sorted.length - 1) {
return sorted[i];
}
} else {
if (count % 2 !== 0) {
return temp;
}
count = 1;
temp = sorted[i];
}
}
}
答案 8 :(得分：0)
如果我们确定只有一个数字会出现奇数次，我们可以在n个比较中对这些数字进行异或运算，找到出现奇数次的数字。如果两位不同，则两位的异或为1，否则为0。真相表如下。
A B A^B
0 0 0
0 1 1
1 0 1
1 1 0
因此，当我们对所有数字进行XOR运算时，最终数字将是出现奇数次的数字。 让我们取一个数字并将其与相同的数字进行异或运算（出现两次）。结果将为0，因为所有位都将相同。现在让我们对相同数字的结果进行异或。现在结果将是该数字，因为先前结果的所有位均为0，并且仅将相同数字的设置位设置为结果。现在将其扩展为n个数字的数组，数字出现偶数次将得到结果0。数字的奇数出现会导致最终结果中的那个数字。
func oddInt(numbers: [Int]) -> Int {
var result = 0
for aNumber in numbers {
result = result ^ aNumber
}
return result
}
答案 9 :(得分：0)
这是O（N）或O（N * log（N））的解决方案
function findOdd(A) {
var count = {};
for (var i = 0; i < A.length; i++) {
var num = A[i];
if (count[num]) {
count[num] = count[num] + 1;
} else {
count[num] = 1;
}
}
var r = 0;
for (var prop in count) {
if (count[prop] % 2 != 0) {
r = prop;
}
}
return parseInt(r); // since object properies are strings
}
答案 10 :(得分：0)
function oddInt(array) {
let result = 0;
for (let element of array) {
result ^= element
}
return result
}
答案 11 :(得分：0)
var oddNumberTimes = (arr) => {
let hashMap = {};
for (let i = 0; i < arr.length; i++) {
hashMap[arr[i]] = hashMap[arr[i]] + 1 || 1;
}
for (let el in hashMap) {
if (hashMap[el] % 2 !== 0) {
return el;
}
}
return -1;
};
答案 12 :(得分：0)
#using python
a=array('i',[1,1,2,3,3])
ans=0
for i in a:
ans^=i
print('The element that occurs odd number of times:',ans)
列表项 输出：
出现奇数次的元素：2
Xor（^）运算符，当奇数为1时，我们可以在输出中得到1
请参阅异或表：
https://www.electronicshub.org/wp-content/uploads/2015/07/TRUTH-TABLE-1.jpg
答案 13 :(得分：0)
您可以使用按位异或：
function oddInt(array) {
return array.reduce(function(c, v) {
return c ^ v;
}, 0);
}
console.log(oddInt([20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5]) == 5);
console.log(oddInt([1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1]) == 10);答案 14 :(得分：-1)
var arr = [1,2,2,2,2,3,4,3,3,3,4,5,5,9,9,10];
var arr1 = [];
for（let i = 0; i
{
var count=0;
for(let j=0;j<arr.length;j++)
{
if(arr[i]==arr[j])
{
count++;
}
}
if(count%2 != 0 )
{
arr1.push(arr[i]);
}
}
的console.log（ARR1）;