LeetCode 564. Find the Closest Palindrome （构造）

class Solution {
public:
    string nearestPalindromic(string n) {
        typedef long long ll;
        if (n == "10" || n == "11") return "9";
        
        int l = n.size();
        int half = (l + 1) / 2;
        string left = n.substr(0, half);
        string right = n.substr(half);
        
        // 情况1: 直接镜像翻转
        string ans1 = (l & 1) ? left + rev_str(left).substr(1) : left + rev_str(left);
        // 如果和原数字相同则不可用
        ll diff1 = ans1 == n ? stoll(n) : abs(stoll(ans1) - stoll(n));
        
        // 情况2: -1
        string left_sub_1 = to_string(stoll(left) - 1);
        string rleft_sub_1 = rev_str(left_sub_1);
        string ans2;
        if (left_sub_1.size() < half) {
            ans2 = (l & 1)
                ? left_sub_1 + rleft_sub_1
                : left_sub_1 + "9" + rleft_sub_1;
        }
        else {
            ans2 = (l & 1)
                ? left_sub_1 + rleft_sub_1.substr(1)
                : left_sub_1 + rleft_sub_1;
        }
        ll diff2 = abs(stoll(ans2) - stoll(n));
        
        // 情况3: +1
        string left_add_1 = to_string(stoll(left) + 1);
        string rleft_add_1 = rev_str(left_add_1);
        string ans3;
        if (left_add_1.size() > half) {
            ans3 = (l & 1) 
                ? left_add_1 + rleft_add_1.substr(2) 
                : left_add_1 + rleft_add_1.substr(1);
        }
        else {
            ans3 = (l & 1)
                ? left_add_1 + rleft_add_1.substr(1)
                : left_add_1 + rleft_add_1;
        }
        ll diff3 = abs(stoll(ans3) - stoll(n));
        
        if (diff2 <= diff1 && diff2 <= diff3) return ans2;
        if (diff1 <= diff2 && diff1 <= diff3) return ans1;
        return ans3;
    }
    
    string rev_str(string a) {
        string b(a);
        reverse(b.begin(), b.end());
        return b;
    }
};