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问题:
We are required in our assignment to find the second smallest integer in one array recursively. However, for the sake of understanding the subject more, I want to do it iteratively first (with the help of this website) and recursively on my own.
Unfortunately, doing it iteratively is quite confusing. I understand that the solution is simple but i can't wrap my head around it.
Below is my code, so far:
public static void main(String[] args)
{
int[] elements = {0 , 2 , 10 , 3, -3 };
int smallest = 0;
int secondSmallest = 0;
for (int i = 0; i < elements.length; i++)
{
for (int j = 0; j < elements.length; j++)
{
if (elements[i] < smallest)
{
smallest = elements[i];
if (elements[j] < secondSmallest)
{
secondSmallest = elements[j];
}
}
}
}
System.out.println("The smallest element is: " + smallest + "\n"+ "The second smallest element is: " + secondSmallest);
}
This works for a few numbers, but not all. The numbers change around because the inner if condition isn't as efficient as the outer if condition.
Array rearrangements are forbidden.
回答1:
Try this one. Second condition is used to catch an event when the smallest number is the first
int[] elements = {-5, -4, 0, 2, 10, 3, -3};
int smallest = Integer.MAX_VALUE;
int secondSmallest = Integer.MAX_VALUE;
for (int i = 0; i < elements.length; i++) {
if(elements[i]==smallest){
secondSmallest=smallest;
} else if (elements[i] < smallest) {
secondSmallest = smallest;
smallest = elements[i];
} else if (elements[i] < secondSmallest) {
secondSmallest = elements[i];
}
}
UPD by @Axel
int[] elements = {-5, -4, 0, 2, 10, 3, -3};
int smallest = Integer.MAX_VALUE;
int secondSmallest = Integer.MAX_VALUE;
for (int i = 0; i < elements.length; i++) {
if (elements[i] < smallest) {
secondSmallest = smallest;
smallest = elements[i];
} else if (elements[i] < secondSmallest) {
secondSmallest = elements[i];
}
}
回答2:
int[] arr = { 4, 1, 2, 0, 6, 1, 2, 0 };
int smallest = Integer.MAX_VALUE;
int smaller = Integer.MAX_VALUE;
int i = 0;
if (arr.length > 2) {
for (i = 0; i < arr.length; i++) {
if (arr[i] < smallest) {
smaller = smallest;
smallest = arr[i];
} else if (arr[i] < smaller && arr[i] > smallest) {
smaller = arr[i];
}
}
System.out.println("Smallest number is " + smallest);
System.out.println("Smaller number is " + smaller);
} else {
System.out.println("Invalid array !");
}
}
回答3:
You can do it in O(n) time. Below is the python code
def second_small(A):
if len(A)<2:
print 'Invalid Array...'
return
small = A[0]
second_small = [1]
if small > A[1]:
second_small,small = A[0],A[1]
for i in range(2,len(A)):
if A[i] < second_small and A[i]!=small:
if A[i] < small:
second_small = small
small = A[i]
else:
second_small = A[i]
print small, second_small
A = [12, 13, 1, 10, 34, 1]
second_small(A)
回答4:
public static int findSecondSmallest(int[] elements) {
if (elements == null || elements.length < 2) {
throw new IllegalArgumentException();
}
int smallest = elements[0];
int secondSmallest = elements[0];
for (int i = 1; i < elements.length; i++) {
if (elements[i] < smallest) {
secondSmallest = smallest;
smallest = elements[i];
}
else if (elements[i] < secondSmallest) {
secondSmallest = elements[i];
}
}
return secondSmallest;
}
回答5:
Simply, you can do this
int[] arr = new int[]{34, 45, 21, 12, 54, 67, 15};
Arrays.sort(arr);
System.out.println(arr[1]);
回答6:
Try this one.
public static void main(String args[]){
int[] array = new int[]{10, 30, 15, 8, 20, 4};
int min, secondMin;
if (array[0] > array[1]){
min = array[1];
secondMin = array[0];
}
else{
min = array[0];
secondMin = array[1];
}
for (int i=2; i<array.length; i++){
if (array[i] < min){
secondMin = min;
min = array[i];
}
else if ((array[i] > min) && (array[i] < secondMin)){
secondMin = array[i];
}
}
System.out.println(secondMin);
}
回答7:
I've used Sort function in javascript
function sumTwoSmallestNumbers(numbers){
numbers = numbers.sort(function(a, b){return a - b; });
return numbers[0] + numbers[1];
};
by providing a compareFunction for the sort functionality array elements are sorted according to the return value of the function.
回答8:
How about this?
int[] result = Arrays.asList(-3, 4,-1,-2).stream()
.reduce(new int[]{Integer.MIN_VALUE, Integer.MIN_VALUE},
(maxValues, x) -> {
if (x > maxValues[0]) {
maxValues[1] = maxValues[0]; //max becomes second max
maxValues[0] = x;
}
else if (x > maxValues[1]) maxValues[1] = x;
return maxValues;
}
, (x, y) -> x);
回答9:
class A{
public static void main (String args[]){
int array[]= {-5, -4, 0, 2, 10, 3, -3};
int min;
int second_min;
if(array[0]<array[1]){
min=array[0];
second_min=array[1];
}else{
min=array[1];
second_min=array[0];
}
for(int i=2;i<array.length;i++){
if(second_min > array[i] && min > array[i]){
second_min=min;
min=array[i];
}else if(second_min > array[i] && min < array[i]){
min=min;
second_min=array[i];
}
}
System.out.println(min);
System.out.println(second_min);
}
}
回答10:
Find the second minimum element of an array in Python, short and simple
def second_minimum(arr):
second = arr[1]
first = arr[0]
for n in arr:
if n < first:
first = n
if n > first and n < second :
second = n
return second
print(second_minimum([-2, 4, 5, -1, 2, 3, 0, -4, 1, 99, -6, -5, -19]))
回答11:
public static void main(String[] args)
{
int[] elements = {-4 , 2 , 10 , -2, -3 };
int smallest = Integer.MAX_VALUE;
int secondSmallest = Integer.MAX_VALUE;
for (int i = 0; i < elements.length; i++)
{
if (smallest>elements[i])
smallest=elements[i];
}
for (int i = 0; i < elements.length; i++)
{
if (secondSmallest>elements[i] && elements[i]>smallest)
secondSmallest=elements[i];
}
System.out.println("The smallest element is: " + smallest + "\n"+ "The second smallest element is: " + secondSmallest);
}
回答12:
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter array size = ");
int size=in.nextInt();
int[] n = new int[size];
System.out.println("Enter "+ size +" values ");
for(int i=0;i<n.length;i++)
n[i] = in.nextInt();
int small=n[0],ssmall=n[0];
// finding small and second small
for(int i=0;i<n.length;i++){
if(small>n[i]){
ssmall=small;
small=n[i];
}else if(ssmall>n[i])
ssmall=n[i];
}
// finding second small if first element itself small
if(small==n[0]){
ssmall=n[1];
for(int i=1;i<n.length;i++){
if(ssmall>n[i]){
ssmall=n[i];
}
}
}
System.out.println("Small "+ small+" sSmall "+ ssmall);
in.close();
}
回答13:
public static void main(String[] args) {
int arr[] = {6,1,37,-4,12,46,5,64,21,2,-4,-3};
int lowest =arr[0];
int sec_lowest =arr[0];
for(int n : arr){
if (lowest > n)
{
sec_lowest = lowest;
lowest = n;
}
else if (sec_lowest > n && lowest != n)
sec_lowest = n;
}
System.out.println(lowest+" "+sec_lowest);
}
回答14:
public class SecondSmallestNumberInArray
{
public static void main(String[] args)
{
int arr[] = { 99, 76, 47, 85, 929, 52, 48, 36, 66, 81, 9 };
int smallest = arr[0];
int secondSmallest = arr[0];
System.out.println("The given array is:");
boolean find = false;
boolean flag = true;
for (int i = 0; i < arr.length; i++)
{
System.out.print(arr[i] + " ");
}
System.out.println("");
while (flag)
{
for (int i = 0; i < arr.length; i++)
{
if (arr[i] < smallest)
{
find = true;
secondSmallest = smallest;
smallest = arr[i];
} else if (arr[i] < secondSmallest) {
find = true;
secondSmallest = arr[i];
}
}
if (find) {
System.out.println("\nSecond Smallest number is Array : -> " + secondSmallest);
flag = false;
} else {
smallest = arr[1];
secondSmallest = arr[1];
}
}
}
}
**Output is**
D:\Java>java SecondSmallestNumberInArray
The given array is:
99 76 47 85 929 52 48 36 66 81 9
Second Smallest number is Array : -> 36
D:\Java>
回答15:
public static int getSecondSmallest(int[] arr){
int smallest = Integer.MAX_VALUE;
int secondSmallest = Integer.MAX_VALUE;
for(int i=0;i<arr.length;i++){
if(smallest > arr[i]){
secondSmallest = smallest;
smallest = arr[i];
}else if (secondSmallest > arr[i] && arr[i] != smallest){
secondSmallest = arr[i];
}
System.out.println(i+" "+smallest+" "+secondSmallest);
}
return secondSmallest;
}
Just gave it a try with some of the test cases and it worked. Please check if it is correct!
回答16:
Try this ...
First condition checks if both values are less than value in array.
Second condition if value is less than small than smallest=element[i]
else secondSmallest=elements[i]..
public static void main(String[] args)
{
int[] elements = {0 , 2 , 10 , 3, -3 };
int smallest = elements[0];
int secondSmallest = 0;
for (int i = 0; i < elements.Length; i++)
{
if (elements[i]<smallest || elements[i]<secondSmallest )
{
if (elements[i] < smallest )
{
secondSmallest = smallest ;
smallest = elements[i];
}
else
{
secondSmallest = elements[i];
}
}
}
System.out.println("The smallest element is: " + smallest + "\n"+ "The second smallest element is: " + secondSmallest);
}
回答17:
Try this, program gives solution for both lowest value and second lowest value of array.
Initialize min and second_min with first element of array.Find out the min value and compare it with second_min value . If it (second_min) is greater than current element of array and min value then the second_min value replace with current element of array.
In case arr[]={2,6,12,15,11,0,3} like this , temp variable used to store previous second_min value.
public class Main
{
public static void main(String[] args) {
//test cases.
int arr[]={6,12,1,11,0};
//int arr[]={0,2,10,3,-3};
//int arr[]={0,0,10,3,-3};
//int arr[]={0,2 ,10, 3,-3};
//int arr[]={12,13,1,10,34,1};
//int arr[]={2,6,12,15,11,0,3};
//int arr[]={2,6,12,15,1,0,3};
//int arr[]={2,6,12,15};
//int arr[]={0,1};
//int arr[]={6,16};
//int arr[]={12};
//int arr[]={6,6,6,6,6,6};
int position_min=0;
int min=arr[0];int second_min=arr[0]; int temp=arr[0];
if(arr.length==1)
{
System.out.println("Lowest value is "+arr[0]+"\n Array length should be greater than 1. ");
}
else if(arr.length==2)
{
if(arr[0]>arr[1])
{
min=arr[1];
second_min=arr[0];
position_min=1;
}
else
{
min=arr[0];
second_min=arr[1];
position_min=0;
}
System.out.println("Lowest value is "+min+"\nSecond lowest value is "+second_min);
}
else
{
for( int i=1;i<arr.length;i++)
{
if(min>arr[i])
{
min=arr[i];
position_min=i;
}
}
System.out.println("Lowest value is "+min);
for(int i=1;i<arr.length;i++)
{
if(position_min==i)
{
}
else
{
if(second_min > min & second_min>arr[i])
{
temp=second_min;
second_min=arr[i];
}
else if(second_min == min )
{
second_min=arr[i];
}
}
}
if(second_min==min )
{
second_min=temp;
}
//just for message if in case all elements are same in array.
if(temp==min && second_min==min)
{
System.out.println("There is no Second lowest element in array.");
}
else{
System.out.println("\nSecond lowest value is "+second_min);
}
}
}
}
回答18:
Here's a Swift version that runs in linear time. Basically, find the smallest number. Then assign the 2nd minimum number as the largest value. Then loop through through the array and find a number greater than the smallest one but also smaller than the 2nd smallest found so far.
func findSecondMinimumElementLinear(in nums: [Int]) -> Int? {
// If the size is less than 2, then returl nil.
guard nums.count > 1 else { return nil }
// First, convert it into a set to reduce duplicates.
let uniqueNums = Array(Set(nums))
// There is no point in sorting if all the elements were the same since it will only leave 1 element
// after the set removed duplicates.
if uniqueNums.count == 1 { return nil }
let min: Int = uniqueNums.min() ?? 0 // O(n)
var secondMinNum: Int = uniqueNums.max() ?? 0 // O(n)
// O(n)
for num in uniqueNums {
if num > min && num < secondMinNum {
secondMinNum = num
}
}
return secondMinNum
}
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