我很难理解在数组中查找第二大数字的方法背后的逻辑。使用的方法是在数组中找到最高但小于之前的最高（已经找到）。我仍然无法弄清楚为什么|| highest_score == second_highest 是必要的。例如我输入了三个数字：98、56、3。没有它，最高和第二高的都是98。请解释一下。
int second highest = score[0];
if (score[i] > second_highest && score[i] < highest_score || highest_score == second_highest)
second_highest = score[i];
【问题讨论】：
second highest 是second_highest 吗？有循环吗？
想法是取两个变量 firstmax fm 和 secondmax sm 并遍历数组。
def print2largest(self,A,N):
#code here
if all(A[x]==A[x-1] for x in range(1,N)):
return -1
sm=0
fs=0
for i in range(N):
if A[i]>fs:
sm=fs
fs=A[i]
elif A[i]>sm and A[i]!=fs:
sm=A[i]
return sm
【讨论】：
let array = [0,12,74,26,82,176,189,8,55,3,189];
let highest=0 ;
let secondHighest = 0;
for (let i = 0; i < array.length; i++) {
if (array[i] > highest) {
// ...shift the current highest number to second highest
secondHighest = highest;
// ...and set the new highest.
highest = array[i];
} else if (highest > array[i] > secondHighest) {
// Just replace the second highest
secondHighest = array[i];
}
}
console.log(secondHighest);
【讨论】：
private static void returnSecondHighetsNumber(int[] a) {
int firstHighest=0;
int secondHighest=0;
if(a[0]>=a[1]) {
firstHighest=a[0];
secondHighest=a[1];
}else {
firstHighest=a[1];
secondHighest=a[0];
}
for (int i = 2; i < a.length; i++) {
if(a[i] >=firstHighest) {
secondHighest=firstHighest;
firstHighest=a[i];
}
else if(a[i]>=secondHighest) {
secondHighest=a[i];
}
}
System.out.println(firstHighest+"---"+secondHighest);
}
【讨论】：
这个 java 程序帮助在任何给定的数组中找到第二大的数字。 首先，我们必须获得最高价值，然后我们必须获得第二名。
class Demo{
public static void main(String args[]){
int arr[]={321,321,321,43,65,234,56,-1,4,45,-67};
int max=arr[0];
int len=arr.length;
for(int i=0;i<len;i++){
if(max<arr[i]){
max=arr[i];
}
}
System.out.println("Max value is "+max);
int secMax=arr[0];
for(int j=0;j<len;j++){
if(max>arr[j]){
if(secMax<arr[j]){
secMax=arr[j];
}
else if(secMax>arr[j]){
secMax=arr[j];
}
}
else if(max==arr[j]){
//
}
}
System.out.println("Sec Max value is "+secMax);
}
}
【讨论】：
数组中的第二大元素： 在 Java 中：
class test2{
public static void main(String[] args) {
int a[] = {1,2,3,9,5,7,6,4,8};
Arrays.sort(a);
int aa = a[a.length -2 ];
System.out.println(aa);
}//main
}//end
在 Python 中：
a = [1, 2, 3, 9, 5, 7, 6, 4, 8]
aa = sorted(list(a))
print(aa)
aaa = aa[-2]
print(aaa)
【讨论】：
最简单的方法是——
public class SecondLargest {
public static void main(String[] args) {
int[] arr = { 1, 2, 5, 6, 3 };
int first = Integer.MIN_VALUE;
int second = Integer.MIN_VALUE;
for (int i = 0; i < arr.length; i++) {
// If current element is smaller than first then update both first
// and second
if (arr[i] > first) {
second = first;
first = arr[i];
}
// If arr[i] is in between first and second then update second
else if (arr[i] > second && arr[i] != first) {
second = arr[i];
}
}
}
}
【讨论】：
我的想法是你假设数组的第一个和第二个成员是你的第一个最大值和第二个最大值。然后，您获取数组的每个新成员并将其与第二个最大值进行比较。不要忘记将第二个最大值与第一个最大值进行比较。如果它更大，只需交换它们。
public static int getMax22(int[] arr){
int max1 = arr[0];
int max2 = arr[1];
for (int i = 2; i < arr.length; i++){
if (arr[i] > max2)
{
max2 = arr[i];
}
if (max2 > max1)
{
int temp = max1;
max1 = max2;
max2 = temp;
}
}
return max2;
}
【讨论】：
导入 java.util.Scanner;
公共类 SecondHighestFromArrayTest {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter size of Array");
int size = scan.nextInt();
int[] arr = new int[size];
for (int i = 0; i < size; i++) {
arr[i] = scan.nextInt();
}
System.out.println("second highest element " + getSecondHighest(arr));
}
public static int getSecondHighest(int arr[]) {
int firstHighest = arr[0];
int secondHighest = arr[0];
for (int i = 0; i < arr.length; i++) {
if (arr[i] > firstHighest) {
secondHighest = firstHighest;
firstHighest = arr[i];
} else if (arr[i] > secondHighest) {
secondHighest = arr[i];
}
}
return secondHighest;
}
}
【讨论】：
我有最简单的逻辑来找到第二大的数字可能是，不是。 逻辑在数组中找到两个数之和，其中具有最高值，然后检查两个简单数中哪个更大............
int ar[]={611,4,556,107,5,55,811};
int sum=ar[0]+ar[1];
int temp=0;
int m=ar[0];
int n=ar[1];
for(int i=0;i<ar.length;i++){
for(int j=i;j<ar.length;j++){
if(i!=j){
temp=ar[i]+ar[j];
if(temp>sum){
sum=temp;
m=ar[i];
n=ar[j];
}
temp=0;
}
}
}
if(m>n){
System.out.println(n);
}
else{
System.out.println(m);
}
【讨论】：
/* Function to print the second largest elements */
void print2largest(int arr[], int arr_size)
{
int i, first, second;
/* There should be atleast two elements */
if (arr_size < 2)
{
printf(" Invalid Input ");
return;
}
first = second = INT_MIN;
for (i = 0; i < arr_size ; i ++)
{
/* If current element is smaller than first
then update both first and second */
if (arr[i] > first)
{
second = first;
first = arr[i];
}
/* If arr[i] is in between first and
second then update second */
else if (arr[i] > second && arr[i] != first)
second = arr[i];
}
if (second == INT_MIN)
printf("There is no second largest elementn");
else
printf("The second largest element is %dn", second);
}
【讨论】：
public class SecondLargestNumber
{
public static void main(String[] args)
{
int[] var={-11,-11,-11,-11,115,-11,-9};
int largest = 0;
int secLargest = 0;
if(var.length == 1)
{
largest = var[0];
secLargest = var[0];
}
else if(var.length > 1)
{
largest= var[0];
secLargest = var[1];
for(int i=1;i<var.length;i++)
{
if(secLargest!=largest)
{
if(var[i]>largest)
{
secLargest = largest;
largest = var[i];
}
else if(var[i]>secLargest && var[i] != largest)
{
secLargest= var[i];
}
}
else
{
if(var[i]>largest)
{
secLargest = largest;
largest = var[i];
}
else
{
secLargest = var[i];
}
}
}
}
System.out.println("Largest: "+largest+" Second Largest: "+secLargest);
}
}
【讨论】：
public void findMax(int a[]) {
int large = Integer.MIN_VALUE;
int secondLarge = Integer.MIN_VALUE;
for (int i = 0; i < a.length; i++) {
if (large < a[i]) {
secondLarge = large;
large = a[i];
} else if (a[i] > secondLarge) {
if (a[i] != large) {
secondLarge = a[i];
}
}
}
System.out.println("Large number " + large + " Second Large number " + secondLarge);
}
上面的代码已经用具有重复条目、负值的整数数组进行了测试。一次检索最大数和第二大数。仅当数组仅包含相同数字的多个副本（例如 {8,8,8,8}）或只有一个数字时，此代码才会失败。
【讨论】：
import java.util.Scanner;
public class SecondLargest {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter size of array : ");
int n = sc.nextInt();
int ar[] = new int[n];
for(int i=0;i<n;i++)
{
System.out.print("Enter value for array : ");
ar[i] = sc.nextInt();
}
int m=ar[0],m2=ar[0];
for(int i=0;i<n;i++)
{
if(ar[i]>m)
m=ar[i];
}
for(int i=0;i<n;i++)
{
if(ar[i]>m2 && ar[i]<m)
m2=ar[i];
}
System.out.println("Second largest : "+m2);
sc.close();
}
}
【讨论】：
使用以下功能
`
public static int secHigh(int arr[]){
int firstHigh = 0,secHigh = 0;
for(int x: arr){
if(x > firstHigh){
secHigh = firstHigh;
firstHigh = x;
}else if(x > secHigh){
secHigh = x;
}
}
return secHigh;
}
函数调用
int secondHigh = secHigh(arr);
【讨论】：
public class SecondHighest {
public static void main(String[] args) {
// TODO Auto-generated method stub
/*
* Find the second largest int item in an unsorted array.
* This solution assumes we have atleast two elements in the array
* SOLVED! - Order N.
* Other possible solution is to solve with Array.sort and get n-2 element.
* However, Big(O) time NlgN
*/
int[] nums = new int[]{1,2,4,3,5,8,55,76,90,34,91};
int highest,cur, secondHighest = -1;
int arrayLength = nums.length;
highest = nums[1] > nums[0] ? nums[1] : nums[0];
secondHighest = nums[1] < nums[0] ? nums[1] : nums[0];
if (arrayLength == 2) {
System.out.println(secondHighest);
} else {
for (int x = 0; x < nums.length; x++) {
cur = nums[x];
int tmp;
if (cur < highest && cur > secondHighest)
secondHighest = cur;
else if (cur > secondHighest && cur > highest) {
tmp = highest;
highest = cur;
secondHighest = tmp;
}
}
System.out.println(secondHighest);
}
}
}
【讨论】：
public static void main(String[] args) {
int[] arr = {0,12,74,56,2,63,45};
int f1 = 1, f2 = 0, temp = 0;
int num = 0;
for (int i = 0; i < arr.length; i++){
num = arr[i];
if (f1 < num) {
temp = f1;
f1 = num;
num = temp;
}
if (f2 < num) {
temp = f2;
f2 = num;
num = temp;
}
}
System.out.println("First Highest " + f1 + " Second Highest " + f2 + " Third " + num);
}
【讨论】：
int arr[] = {25, 5, 35, 26, 6, 270, 0};
int large1 = arr[0];
int large2 = arr[1];
for (int i = 2; i < arr.length; i++) {
if (arr[i] > large1) {
if (large1 > large2)
large2 = large1;
large1 = arr[i];
} else if (arr[i] > large2)
large2 = arr[i];
}
System.out.println("Large1 " + large1);
System.out.println("Large2 " + large2);
【讨论】：
求第二大数：
public class SecondMaxNumbers {
public void printTwoMaxNumbers(int[] nums){
int maxOne = 0;
int maxTwo = 0;
for(int n:nums){
if(maxOne < n){
maxTwo = maxOne;
maxOne =n;
} else if(maxTwo < n){
maxTwo = n;
}
}
System.out.println("Second Max Number: "+maxTwo);
}
public static void main(String a[]){
int num[] = {10,20,30,40,50,60,70};
SecondMaxNumbers sec = new SecondMaxNumbers();
tmn.printTwoMaxNumbers(num);
}
}
【讨论】：
O(n/2) 中的第二大
public class SecMaxNum {
// second Largest number with O(n/2)
/**
* @author Rohan Kamat
* @Date Feb 04, 2016
*/
public static void main(String[] args) {
int[] input = { 1, 5, 10, 11, 11, 4, 2, 8, 1, 8, 9, 8 };
int large = 0, second = 0;
for (int i = 0; i < input.length - 1; i = i + 2) {
// System.out.println(i);
int fist = input[i];
int sec = input[i + 1];
if (sec >= fist) {
int temp = fist;
fist = sec;
sec = temp;
}
if (fist >= second) {
if (fist >= large) {
large = fist;
} else {
second = fist;
}
}
if (sec >= second) {
if (sec >= large) {
large = sec;
} else {
second = sec;
}
}
}
}
}
【讨论】：
公共类 SecondHighInIntArray {
public static void main(String[] args) {
int[] intArray=new int[]{2,2,1};
//{2,2,1,12,3,7,9,-1,-5,7};
int secHigh=findSecHigh(intArray);
System.out.println(secHigh);
}
private static int findSecHigh(int[] intArray) {
int highest=Integer.MIN_VALUE;
int sechighest=Integer.MIN_VALUE;
int len=intArray.length;
for(int i=0;i<len;i++)
{
if(intArray[i]>highest)
{
sechighest=highest;
highest=intArray[i];
continue;
}
if(intArray[i]<highest && intArray[i]>sechighest)
{
sechighest=intArray[i];
continue;
}
}
return sechighest;
}
}
【讨论】：
请试试这个：使用此方法，您可以对数组中的第二大数进行罚款，即使数组包含随机数。第一个循环用于解决数组的第一个索引最大的问题。
public class secondLargestnum {
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] array = new int[6];
array[0] = 10;
array[1] = 80;
array[2] = 5;
array[3] = 6;
array[4] = 50;
array[5] = 60;
int tem = 0;
for (int i = 0; i < array.length; i++) {
if (array[0]>array[i]) {
tem = array[0];
array[0] = array[array.length-1];
array[array.length-1] = tem;
}
}
Integer largest = array[0];
Integer second_largest = array[0];
for (int i = 0; i < array.length; i++) {
if (largest<array[i]) {
second_large = largest;
largest = array[i];
}
else if (second_large<array[i]) {
second_large = array[i];
}
}
System.out.println("largest number "+largest+" and second largest number "+second_largest);
}
}
【讨论】：
private static int SecondBiggest(int[] vector)
{
if (vector == null)
{
throw new ArgumentNullException("vector");
}
if (vector.Length < 2)
{
return int.MinValue;
}
int max1 = vector[0];
int max2 = vector[1];
for (int i = 2; i < vector.Length; ++i)
{
if (max1 > max2 && max1 != vector[i])
{
max2 = Math.Max(max2, vector[i]);
}
else if (max2 != vector[i])
{
max1 = Math.Max(max1, vector[i]);
}
}
return Math.Min(max1, max2);
}
这会将重复的数字视为相同的数字。如果您希望所有最大的和第二大的都重复，您可以更改条件检查。
【讨论】：
问题： 问题是要获得第二大数组元素。
观察： 第二大数定义为从数组中的最大元素中减去后具有最小差异的数。
解决方案： 这是一个两通解决方案。第一遍是找到最大数量。第二遍是找到与其他数组元素相比与最大元素差异最小的元素。示例：在数组 [2, 3, 6, 6, 5] 最大值 = 6 和第二最大值 = 5 中，因为它与最大元素 6 - 5 = 1 的差异最小，所以第二大的解 = 5
function printSecondMax(myArray) {
var x, max = myArray[0];
// Find maximum element
for(x in myArray){
if(max < myArray[x]){
max = myArray[x];
}
}
var secondMax = myArray[0], diff = max - secondMax;
// Find second max, an element that has min diff with the max
for(x in myArray){
if(diff != 0 && (max - myArray[x]) != 0 && diff > (max - myArray[x])){
secondMax = myArray[x];
diff = max - secondMax;
}
}
console.log(secondMax);
}
复杂度：O(n)，这是最简单的方法。
为了更有效地找到最大元素，可以查看max heap，调用 max-heapify 将花费 O(log n) 时间来找到最大值，然后弹出顶部元素给出最大值。要获得第二个最大值，请在弹出顶部后进行 max-heapify 并继续弹出直到您得到一个小于最大值的数字。这将是第二个最大值。该解决方案具有 O(n log n) 复杂度。
【讨论】：
如果这个问题来自面试官，那么请不要使用排序技术或不要使用任何内置方法，如 Arrays.sort 或 Collection.sort。这个问题的目的是您的解决方案在性能方面的优化程度，因此最好的选择就是使用您自己的逻辑和 O(n-1) 实现来实现。以下代码仅供初学者使用，不适合有经验的人。
public void printLargest(){
int num[] ={ 900,90,6,7,5000,4,60000,20,3};
int largest = num[0];
int secondLargest = num[1];
for (int i=1; i<num.length; i++)
{
if(largest < num[i])
{
secondLargest = largest;
largest = num[i];
}
else if(secondLargest < num[i]){
secondLargest = num[i];
}
}
System.out.println("Largest : " +largest);
System.out.println("Second Largest : "+secondLargest);
}
【讨论】：
public class SecondandThirdHighestElement {
public static void main(String[] args) {
int[] arr = {1,1,2,3,8,1,2,3,3,3,2,3,101,6,6,7,8,8,1001,99,1,0};
// create three temp variable and store arr of first element in that temp variable so that it will compare with other element
int firsttemp = arr[0];
int secondtemp = arr[0];
int thirdtemp = arr[0];
//check and find first highest value from array by comparing with other elements if found than save in the first temp variable
for (int i = 0; i < arr.length; i++) {
if(firsttemp <arr[i]){
firsttemp = arr[i];
}//if
}//for
//check and find the second highest variable by comparing with other elements in an array and find the element and that element should be smaller than first element array
for (int i = 0; i < arr.length; i++) {
if(secondtemp < arr[i] && firsttemp>arr[i]){
secondtemp = arr[i];
}//if
}//for
//check and find the third highest variable by comparing with other elements in an array and find the element and that element should be smaller than second element array
for (int i = 0; i < arr.length; i++) {
if(thirdtemp < arr[i] && secondtemp>arr[i]){
thirdtemp = arr[i];
}//if
}//for
System.out.println("First Highest Value:"+firsttemp);
System.out.println("Second Highest Value:"+secondtemp);
System.out.println("Third Highest Value:"+thirdtemp);
}//main
}//class
【讨论】：
我提供的解决方案不在 JAVA 程序中（用 JavaScript 编写），但它需要 o(n/2) 次迭代才能找到最高和第二高的数字。
工作提琴手链接Fiddler link
var num=[1020215,2000,35,2,54546,456,2,2345,24,545,132,5469,25653,0,2315648978523];
var j=num.length-1;
var firstHighest=0,seoncdHighest=0;
num[0] >num[num.length-1]?(firstHighest=num[0],seoncdHighest=num[num.length-1]):(firstHighest=num[num.length-1], seoncdHighest=num[0]);
j--;
for(var i=1;i<=num.length/2;i++,j--)
{
if(num[i] < num[j] )
{
if(firstHighest < num[j]){
seoncdHighest=firstHighest;
firstHighest= num[j];
}
else if(seoncdHighest < num[j] ) {
seoncdHighest= num[j];
}
}
else {
if(firstHighest < num[i])
{
seoncdHighest=firstHighest;
firstHighest= num[i];
}
else if(seoncdHighest < num[i] ) {
seoncdHighest= num[i];
}
}
}
【讨论】：
static int secondLargest(int[] input){
int largest , secondlargest;
if(input[0]>input[1])
{
largest=input[0];
secondlargest = input[1];
}
else
{
secondlargest = input[0];
largest =input[1];
}
for(int i =2 ;i<input.length;i++){
if(input[i]>largest)
{
secondlargest = largest;
largest = input[i];
}
else if(input[i]>secondlargest){
secondlargest = input[i];
}
}
return secondlargest;
}
【讨论】：
如果我们可以使用内置函数，我认为要找到第二个最高不，我们需要这些行
int[] randomIntegers = {1, 5, 4, 2, 8, 1, 1, 6, 7, 8, 9};
Arrays.sort(randomIntegers);
System.out.println(randomIntegers[randomIntegers.length-2]);
【讨论】：
Scanner sc = new Scanner(System.in);
System.out.println("\n number of input sets::");
int value=sc.nextInt();
System.out.println("\n input sets::");
int[] inputset;
inputset = new int[value];
for(int i=0;i<value;i++)
{
inputset[i]=sc.nextInt();
}
int maxvalue=inputset[0];
int secondval=inputset[0];
for(int i=1;i<value;i++)
{
if(inputset[i]>maxvalue)
{
maxvalue=inputset[i];
}
}
for(int i=1;i<value;i++)
{
if(inputset[i]>secondval && inputset[i]<maxvalue)
{
secondval=inputset[i];
}
}
System.out.println("\n maxvalue"+ maxvalue);
System.out.println("\n secondmaxvalue"+ secondval);
【讨论】：
这是我在 C 中的答案，O(N) 复杂度时间。 数组只传递一次，只有三个变量。 该解决方案非常直观且易于理解。
#include <stdio.h>
int second(int arr[],int size)
{
int i, max , secondmax;
if (arr[0] > arr[1]){
max = arr[0];
secondmax = arr[1];
} else {
max = arr[1];
secondmax = arr[0];
}
for (i = 2; i < size; i++)
{
if ((arr[i] < max) && (arr[i] < secondmax)) {
continue;
}
if ((arr[i] < max) && (arr[i] > secondmax)) {
secondmax = arr[i];
continue;
}
if ((arr[i] > max)) {
secondmax = max;
max = arr[i];
continue;
}
}
return secondmax;
}
void main()
{
int arr[] = { 1,10,5,7};
int size = sizeof(arr) / sizeof(arr[0]);
int secondmax = second(arr,size);
printf("%d \n\n", secondmax);
}
【讨论】：