LeetCode 564. Find the Closest Palindrome

原题链接在这里：https://leetcode.com/problems/find-the-closest-palindrome/

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.

题解：

First find all the candidates, add "9999..999", "10000..0001". say n = "10", its closet palindrome is "9".

Add first half, + reverse of (first half). 

e.g. 12567, its first half is 125, the candiates couldbe 12521, 12421, 12621.

Then check all the candidates and find the closest one.

Time Complexity: O(mn). m = is candidates list size. n is candiate length.

Space: O(mn).

AC Java:

class Solution {
    public String nearestPalindromic(String n) {
        if(n == null || n.length() == 0){
            return n;
        }
        
        int len = n.length();
        int ind = (len + 1) / 2;
        long left = Long.valueOf(n.substring(0, ind));
        List<Long> cans = new ArrayList<>();
        boolean isEven = len % 2 == 0;
        cans.add(getPal(left, isEven));
        cans.add(getPal(left + 1, isEven));
        cans.add(getPal(left - 1, isEven));
        cans.add((long)(Math.pow(10, len - 1) - 1));
        cans.add((long)(Math.pow(10, len) + 1));
        
        long diff = Long.MAX_VALUE;
        long res = 0;
        long original = Long.valueOf(n);
        for(long can : cans){
            if(can == original){
                continue;
            }
            
            if(Math.abs(can - original) < diff){
                diff= Math.abs(can - original);
                res = can;
            }else if(Math.abs(can - original) == diff){
                res = Math.min(res, can);
            }
        }
        
        return String.valueOf(res);
    }
    
    private long getPal(long left, boolean isEven){
        long res = left;
        if(!isEven){
            left /= 10;
        }
        
        while(left != 0){
            res = res * 10 + left % 10;
            left /= 10;
        }
        
        return res;
    }
}

类似Next Palindrome Using Same Digits.