564. Find the Closest Palindrome
一道數(shù)學(xué)題。。。重點(diǎn)是找到各種情況并且對(duì)其簡(jiǎn)化。分成兩種情況，第一種是如果S[:half] 的最后一個(gè)值 +（-1，0，1）然后翻轉(zhuǎn)，第二種情況是如果要找的值和S不一樣長(zhǎng)，那么肯定是 "9...9"和“10..01”之間的一個(gè)

class Solution(object):
    def nearestPalindromic(self, n):
        """
        :type n: str
        :rtype: str
        """
        S = n
        K = len(S)
        candidates = [str(10**k + d) for k in (K-1, K) for d in (-1, 1)]
        prefix = S[:(K+1)/2]
        P = int(prefix)
        for start in map(str, (P-1, P, P+1)):
            candidates.append(start + (start[:-1] if K%2 else start)[::-1])

        def delta(x):
            return abs(int(S) - int(x))

        ans = None
        for cand in candidates:
            if cand != S and not cand.startswith('00'):
                if (ans is None or delta(cand) < delta(ans) or
                        delta(cand) == delta(ans) and int(cand) < int(ans)):
                    ans = cand
        return ans