Final answer:
Using nextInt() followed by nextLine() in the Scanner class often leads to the nextLine() method receiving the leftover newline character, causing it to skip over the intended input. A workaround is to use nextLine() after nextInt() to consume the end-of-line character or to parse inputs as strings and convert them manually.
Step-by-step explanation:
When you use the Scanner class nextInt() method followed immediately by the nextLine() method in Java, you might encounter an issue. The nextInt() method reads the integer value but does not read the newline character (\\) at the end of the line. As a result, when nextLine() is called afterwards, it reads the leftover newline character from the input buffer, which leads to skipping the actual input you intended to read with nextLine().
To avoid this, you can call nextLine() after nextInt() to consume the newline character before you actually want to use nextLine() to capture a full line of input, or use only nextLine() coupled with parsing functions to handle all input.