# Easy Java solution with comments

• ``````    public String nearestPalindromic(String n) {
// the length of the left (head) part, and the length of the right (tail) part
int left = (n.length() + 1) / 2, right = n.length() - left;
long origin = Long.parseLong(n); // original number
long diff = Long.MAX_VALUE, num = 0;
/**
* in order to get closest palindrome, we just need to check palindrome
* numbers that can be constructed from: (left - 1), (left + 0) and
* (left + 1).
*/
for (int i = -1; i <= 1; i++) {
long ret = getPalindrome(head + i, right);
if (ret != origin && Math.abs(ret - origin) < diff) {
diff = Math.abs(ret - origin);
num = ret;
}
}
return Long.toString(num);
}

/**
* Construct palindrome from head. Don't need to check if the length of the original
* number is odd or not. The right part must have the same length as the right part of the
* original number.
*/
private long getPalindrome(long head, int rightLen) {
StringBuilder sb = new StringBuilder(str);
/**
* Corner cases to consider:
* 1. numbers like "10" has a left of "1" which will be "0" when checking (head - 1);
* 2. numbers like "1000" has a left of "10" which will be "9" when checking (head - 1),
*    this is the case that left has a shorter length than right;
*
* By checking (head == 0) || rightLen > str.length(), we kill them all.
*/
if (head == 0l || rightLen > str.length()) {
for (int i = 0; i < rightLen; i++)
sb.append('9');
} else {
/**
* Construct the palindrome numbers from head. This is pretty straight forward.
*/
for (int i = rightLen - 1; i >= 0; i--)
sb.append(str.charAt(i));
}
return Long.parseLong(sb.toString());
}
``````

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