0564-find-the-closest-palindrome.java

class Solution {
    public String nearestPalindromic(String n) {
        // edge cases, no
        
        int len = n.length();
        int i = len % 2 == 0 ? len / 2 - 1: len / 2;
        long left = Long.parseLong(n.substring(0, i+1));
        
        // input: n 12345
        List<Long> candidate = new ArrayList<>();
        candidate.add(getPalindrome(left, len % 2 == 0)); // 12321
        candidate.add(getPalindrome(left+1, len % 2 == 0)); // 12421
        candidate.add(getPalindrome(left-1, len % 2 == 0)); // 12221
        candidate.add((long)Math.pow(10, len-1) - 1); // 9999
        candidate.add((long)Math.pow(10, len) + 1); // 100001
        
        long diff = Long.MAX_VALUE, res = 0, nl = Long.parseLong(n);
        for (long cand : candidate) {
            if (cand == nl) continue;
            if (Math.abs(cand - nl) < diff) {
                diff = Math.abs(cand - nl);
                res = cand;
            } else if (Math.abs(cand - nl) == diff) {
                res = Math.min(res, cand);
            }
        }
        
        return String.valueOf(res);
    }
    
    private long getPalindrome(long left, boolean even) {
        long res = left;
        if (!even) left = left / 10;
        while (left > 0) {
            res = res * 10 + left % 10;
            left /= 10;
        }
        return res;
    }
}

/*
get first half, then compare 5 cases: +0(itself not palindrome), +1 / -1 / 9...9 / 10..01 (itself palindrome)
*/