您多次编写正确的Java代码，并根据约束进行了必要的优化，这经常发生。但是，您会获得TLE😢。
发生这种情况的原因是Java使用Scanner类进行输入和输出输出所花费的时间比BufferedReader和StringBuffer类要慢。在此处详细阅读有关扫描仪类的信息。
看一些技巧来摆脱这个TLE问题(当您的逻辑明显正确时)?
Tip 1 : Avoid using Scanner Class and try to use BufferedReader class.
Tip 2 : Try to use StringBuffer class in case you have to print large number of data.
让我们从GeeksforGeeks练习中解决一个问题，并解决TLE问题：
问题：将0、1、2和2的数组隔离
简而言之，问题在于给定一个0、1和2的数组。我们必须隔离数组开头的所有0，数组中间的所有1和数组末尾的所有2。
例子：
Input : 1 1 2 0 0 2 1
Output : 0 0 1 1 1 2 2
方法：分离0、1和2的数组
下面是上述方法的实现：
// Program to segragate the
// array of 0s, 1s and 2s
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG {
public static void main(String[] args)
{
// Using Scanner class to take input
Scanner sc = new Scanner(System.in);
// Number of testcase input
int t = sc.nextInt();
// Iterating through all the testcases
while (t-- > 0) {
// Input n, i.e. size of array
int n = sc.nextInt();
int arr[] = new int[n];
// Taking input of array elements
for (int i = 0; i < n; i++)
arr[i] = sc.nextInt();
// Calling function to segragate
// input array
segragateArr(arr, n);
// printing the modified array
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
}
sc.close();
}
// Function to segragate 0s, 1s and 2s
public static void segragateArr(int arr[], int n)
{
/*
low : to keep left index
high : to keep right index
mid : to get middle element
*/
int low = 0, high = n - 1, mid = 0;
// Iterating through the array and
// segregating elements
while (mid <= high) {
// If element at mid is 0
// move it to left
if (arr[mid] == 0) {
int temp = arr[low];
arr[low] = arr[mid];
arr[mid] = temp;
low++;
mid++;
}
// If element at mid is 1
// nothing to do
else if (arr[mid] == 1) {
mid++;
}
// If element at mid is 2
// move it to last
else {
int temp = arr[mid];
arr[mid] = arr[high];
arr[high] = temp;
high--;
}
}
}
}
根据我们的期望，它应该通过所有测试用例，并在GeeksforGeeks实践中被接受。但是，当我们在GeeksforGeeks IDE上提交此代码时，它将显示TLE。 
这表示我们已经超过了预期的时间限制。没问题，让我们使用上面给出的提示。
- 使用BufferedReader进行输入。
- 使用StringBuffer保存和打印输出。
方法：分离0、1和2的数组
以下是用于分隔0、1和2的Java代码的实现
// Java program to segragate
// array of 0s, 1s and 2s
import java.io.*;
import java.util.*;
class GFG {
// Driver Code
public static void main(String[] args) throws IOException
{
// Using BufferedReader class to take input
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
// taking input of number of testcase
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
// n : size of array
int n = Integer.parseInt(br.readLine());
// Declaring array
int arr[] = new int[n];
// to read multiple integers line
String line = br.readLine();
String[] strs = line.trim().split("\\s+");
// array elements input
for (int i = 0; i < n; i++)
arr[i] = Integer.parseInt(strs[i]);
// Calling Functions to segregate Array elements
segragateArr(arr, n);
// Using string buffer to append each output in a string
StringBuffer sb = new StringBuffer();
for (int i = 0; i < n; i++)
sb.append(arr[i] + " ");
// finally printing the string
System.out.println(sb);
}
br.close();
}
// Function to segragate 0s, 1s and 2s
public static void segragateArr(int arr[], int n)
{
/*
low : to keep left index
high : to keep right index
mid : to get middle element
*/
int low = 0, high = n - 1, mid = 0;
// Iterating through the array and
// segregating elements
while (mid <= high) {
// If element at mid is 0
// move it to left
if (arr[mid] == 0) {
int temp = arr[low];
arr[low] = arr[mid];
arr[mid] = temp;
low++;
mid++;
}
// If element at mid is 1
// nothing to do
else if (arr[mid] == 1) {
mid++;
}
// If element at mid is 2
// move it to last
else {
int temp = arr[mid];
arr[mid] = arr[high];
arr[high] = temp;
high--;
}
}
}
}
伟大的！您已经升级。
Java TLE问题?似乎非常简单:)。您现在可以尝试。
如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。