给定整数 n 和 k,找到 1 到 n 中字典序第 k 小的数字。java
注意:1 ≤ k ≤ n ≤ 109。ide
示例 :spa
输入:
n: 13 k: 2code
输出:
10it
解释:
字典序的排列是 [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9],因此第二小的数字是 10。io
PS:
十叉树ast
class Solution { public int findKthNumber(int n, int k) { int cur = 1; k--; while (k > 0) { long step = 0, first = cur, last = cur + 1; while (first <= n) { step += Math.min(last, (long) (n + 1)) - first; first *= 10; last *= 10; } if (step > k) { //在树里 k--; cur *= 10; } if (step <= k) { //不在树里 k -= step; ++cur; } } return cur; } }