440. K-th Smallest in Lexicographical Order

最新推荐文章于 2023-02-05 09:56:19 发布

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本文介绍了一种高效算法，用于找到从1到n范围内字典序第K小的整数。通过构建一棵10叉树，并利用递归计数方法，解决了传统遍历方法效率低下的问题。

Given integers n and k, find the lexicographically k-th smallest integer in the range from 1 to n.

Note: 1 ≤ k ≤ n ≤ 109.

Example:

Input:
n: 13   k: 2

Output:
10

Explanation:
The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10.

思路一：直接遍历求解，会TLE，程序如下所示：

class Solution {
    public int findKthNumber(int n, int k) {
        int cnt = 0, index = 1;
        for (int i = 1; i <= n; ++ i){
            cnt ++;
            if (cnt == k){
                return index;
            }
            if ((long)index*10 <= (long)n){
                index *= 10;
            }
            else {
                if (index + 1 > n){
                    index = index/10;
                }
                else {
                    if ((index%10) == 9){
                        while ((index%10) == 9){
                            index = index/10;
                        }
                    }
                }
                index ++;
            }
        }
        return cnt;
    }
}

思路2：本例其实可以看成一颗10叉数，链接点击打开，程序如下所示：

class Solution {
    public int getCount(int p, int q, int n){
        int result = 0;
        long a = (long)p, b = (long)q;
        while (a <= n){
            result += Math.min(b, n+1) - a;
            a *= 10;
            b *= 10;
        }
        return result;
    }
    
    public int findKthNumber(int n, int k) {
        int result = 1;
        while (k > 1){
            int count = getCount(result, result + 1, n);
            if (count < k){
                result += 1;
                k -= count;
            }
            else {
                result *= 10;
                k --;
            }
        }
        return result;
    }
}