[Leetcode] 564. Find the Closest Palindrome 解题报告

题目：

Given an integer n, find the closest integer (not including itself), which is a palindrome.

The 'closest' is defined as absolute difference minimized between two integers.

Example 1:

Input: "123"
Output: "121"

Note:

1. The input n is a positive integer represented by string, whose length will not exceed 18.
2. If there is a tie, return the smaller one as answer.

思路：

和n最接近的回文数根据位数划分，则只有3种可能：

1：比n多1位的数，例如对于99而言，距离它最近的回文数就是101；

2：比n少1位的数，例如对于101， 距离它最近的回文数就是99；

3：和n位数一样，并且其后一半是n的前一半的翻转的数。这样的数又可以分为3种：1）中间数和n相同的，例如123的最近回文数就是121；2）中间数比n大1，例如189的最近回文数就是191；3）中间数比n小1，例如920的最近回文数就是919。

所以只要把这5个数分别计算出来，然后比较哪一个和n最接近即可。

代码：

class Solution {
public:
    string nearestPalindromic(string n) {
        int l = n.size();
        set<long> candidates;
        // biggest, one more digit, 10...01
        candidates.insert(long(pow(10, l)) + 1);
        // smallest, one less digit, 9...9 or 0
        candidates.insert(long(pow(10, l - 1)) - 1);
        // the closest must be in middle digit +1, 0, -1, then flip left to right
        long prefix = stol(n.substr(0, (l + 1) / 2));
        for ( long i = -1; i <= 1; ++i ) {
            string p = to_string(prefix + i);
            string pp = p + string(p.rbegin() + (l & 1), p.rend());
            candidates.insert(stol(pp));
        }
        long num = stol(n), minDiff = LONG_MAX, diff, minVal;
        candidates.erase(num);                  // cannot be itself
        for ( long val : candidates ) {
            diff = abs(val - num);
            if ( diff < minDiff ) {
                minDiff = diff;
                minVal = val;
            }
        }
        return to_string(minVal);
    }
};