README.md

comments	difficulty	edit_url	tags
true	困难	https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0440.K-th%20Smallest%20in%20Lexicographical%20Order/README.md	字典树

440. 字典序的第 K 小数字

English Version

题目描述

给定整数 n 和 k，返回  [1, n] 中字典序第 k 小的数字。

 

示例 1:

输入: n = 13, k = 2
输出: 10
解释: 字典序的排列是 [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9]，所以第二小的数字是 10。

示例 2:

输入: n = 1, k = 1
输出: 1

 

提示:

• 1 <= k <= n <= 109

解法

方法一

Python3

class Solution:
    def findKthNumber(self, n: int, k: int) -> int:
        def count(curr):
            next, cnt = curr + 1, 0
            while curr <= n:
                cnt += min(n - curr + 1, next - curr)
                next, curr = next * 10, curr * 10
            return cnt

        curr = 1
        k -= 1
        while k:
            cnt = count(curr)
            if k >= cnt:
                k -= cnt
                curr += 1
            else:
                k -= 1
                curr *= 10
        return curr

Java

class Solution {
    private int n;

    public int findKthNumber(int n, int k) {
        this.n = n;
        long curr = 1;
        --k;
        while (k > 0) {
            int cnt = count(curr);
            if (k >= cnt) {
                k -= cnt;
                ++curr;
            } else {
                --k;
                curr *= 10;
            }
        }
        return (int) curr;
    }

    public int count(long curr) {
        long next = curr + 1;
        long cnt = 0;
        while (curr <= n) {
            cnt += Math.min(n - curr + 1, next - curr);
            next *= 10;
            curr *= 10;
        }
        return (int) cnt;
    }
}

C++

class Solution {
public:
    int n;

    int findKthNumber(int n, int k) {
        this->n = n;
        --k;
        long long curr = 1;
        while (k) {
            int cnt = count(curr);
            if (k >= cnt) {
                k -= cnt;
                ++curr;
            } else {
                --k;
                curr *= 10;
            }
        }
        return (int) curr;
    }

    int count(long long curr) {
        long long next = curr + 1;
        int cnt = 0;
        while (curr <= n) {
            cnt += min(n - curr + 1, next - curr);
            next *= 10;
            curr *= 10;
        }
        return cnt;
    }
};

Go

func findKthNumber(n int, k int) int {
	count := func(curr int) int {
		next := curr + 1
		cnt := 0
		for curr <= n {
			cnt += min(n-curr+1, next-curr)
			next *= 10
			curr *= 10
		}
		return cnt
	}
	curr := 1
	k--
	for k > 0 {
		cnt := count(curr)
		if k >= cnt {
			k -= cnt
			curr++
		} else {
			k--
			curr *= 10
		}
	}
	return curr
}