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564. 寻找最近的回文数

题目描述

给定一个表示整数的字符串 n ，返回与它最近的回文整数（不包括自身）。如果不止一个，返回较小的那个。

“最近的”定义为两个整数差的绝对值最小。

 

示例 1:

输入: n = "123"
输出: "121"

示例 2:

输入: n = "1"
输出: "0"
解释: 0 和 2是最近的回文，但我们返回最小的，也就是 0。

 

提示:

  • 1 <= n.length <= 18
  • n 只由数字组成
  • n 不含前导 0
  • n 代表在 [1, 1018 - 1] 范围内的整数

解法

方法一

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class Solution:
    def nearestPalindromic(self, n: str) -> str:
        x = int(n)
        l = len(n)
        res = {10 ** (l - 1) - 1, 10**l + 1}
        left = int(n[: (l + 1) >> 1])
        for i in range(left - 1, left + 2):
            j = i if l % 2 == 0 else i // 10
            while j:
                i = i * 10 + j % 10
                j //= 10
            res.add(i)
        res.discard(x)

        ans = -1
        for t in res:
            if (
                ans == -1
                or abs(t - x) < abs(ans - x)
                or (abs(t - x) == abs(ans - x) and t < ans)
            ):
                ans = t
        return str(ans)

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class Solution {
    public String nearestPalindromic(String n) {
        long x = Long.parseLong(n);
        long ans = -1;
        for (long t : get(n)) {
            if (ans == -1 || Math.abs(t - x) < Math.abs(ans - x)
                || (Math.abs(t - x) == Math.abs(ans - x) && t < ans)) {
                ans = t;
            }
        }
        return Long.toString(ans);
    }

    private Set<Long> get(String n) {
        int l = n.length();
        Set<Long> res = new HashSet<>();
        res.add((long) Math.pow(10, l - 1) - 1);
        res.add((long) Math.pow(10, l) + 1);
        long left = Long.parseLong(n.substring(0, (l + 1) / 2));
        for (long i = left - 1; i <= left + 1; ++i) {
            StringBuilder sb = new StringBuilder();
            sb.append(i);
            sb.append(new StringBuilder(i + "").reverse().substring(l & 1));
            res.add(Long.parseLong(sb.toString()));
        }
        res.remove(Long.parseLong(n));
        return res;
    }
}

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class Solution {
public:
    string nearestPalindromic(string n) {
        long x = stol(n);
        long ans = -1;
        for (long t : get(n))
            if (ans == -1 || abs(t - x) < abs(ans - x) || (abs(t - x) == abs(ans - x) && t < ans))
                ans = t;
        return to_string(ans);
    }

    unordered_set<long> get(string& n) {
        int l = n.size();
        unordered_set<long> res;
        res.insert((long) pow(10, l - 1) - 1);
        res.insert((long) pow(10, l) + 1);
        long left = stol(n.substr(0, (l + 1) / 2));
        for (long i = left - 1; i <= left + 1; ++i) {
            string prefix = to_string(i);
            string t = prefix + string(prefix.rbegin() + (l & 1), prefix.rend());
            res.insert(stol(t));
        }
        res.erase(stol(n));
        return res;
    }
};

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func nearestPalindromic(n string) string {
    l := len(n)
    res := []int{int(math.Pow10(l-1)) - 1, int(math.Pow10(l)) + 1}
    left, _ := strconv.Atoi(n[:(l+1)/2])
    for _, x := range []int{left - 1, left, left + 1} {
        y := x
        if l&1 == 1 {
            y /= 10
        }
        for ; y > 0; y /= 10 {
            x = x*10 + y%10
        }
        res = append(res, x)
    }
    ans := -1
    x, _ := strconv.Atoi(n)
    for _, t := range res {
        if t != x {
            if ans == -1 || abs(t-x) < abs(ans-x) || abs(t-x) == abs(ans-x) && t < ans {
                ans = t
            }
        }
    }
    return strconv.Itoa(ans)
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

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/**
 * @param {string} n
 * @return {string}
 */

function nearestPalindromic(n) {
    const x = BigInt(n);
    let ans = null;

    for (const t of getCandidates(n)) {
        if (
            ans === null ||
            absDiff(t, x) < absDiff(ans, x) ||
            (absDiff(t, x) === absDiff(ans, x) && t < ans)
        ) {
            ans = t;
        }
    }

    return ans.toString();
}

function getCandidates(n) {
    const length = n.length;
    const res = new Set();

    res.add(BigInt(Math.pow(10, length - 1) - 1));
    res.add(BigInt(Math.pow(10, length) + 1));

    const left = BigInt(n.substring(0, Math.ceil(length / 2)));

    for (let i = left - 1n; i <= left + 1n; i++) {
        const prefix = i.toString();
        const t =
            prefix +
            prefix
                .split('')
                .reverse()
                .slice(length % 2)
                .join('');
        res.add(BigInt(t));
    }

    res.delete(BigInt(n));
    return res;
}

function absDiff(a, b) {
    return a > b ? a - b : b - a;
}

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