76 lines
1.8 KiB
Java
76 lines
1.8 KiB
Java
//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
|
||
//
|
||
// Related Topics 字典树 👍 389 👎 0
|
||
|
||
package leetcode.editor.cn;
|
||
|
||
//440:字典序的第K小数字
|
||
public class KThSmallestInLexicographicalOrder {
|
||
public static void main(String[] args) {
|
||
Solution solution = new KThSmallestInLexicographicalOrder().new Solution();
|
||
// TO TEST
|
||
System.out.println(solution.findKthNumber(13, 2));
|
||
System.out.println(solution.findKthNumber(1, 1));
|
||
System.out.println(solution.findKthNumber(100, 90));
|
||
System.out.println(solution.findKthNumber(10000, 10000));
|
||
}
|
||
|
||
//leetcode submit region begin(Prohibit modification and deletion)
|
||
class Solution {
|
||
public int findKthNumber(int n, int k) {
|
||
for (int i = 1; i <= n; i++) {
|
||
if (k == 1) {
|
||
return i;
|
||
}
|
||
int count = getCount(i, n);
|
||
if (count < k) {
|
||
k -= count;
|
||
} else {
|
||
i = i * 10 - 1;
|
||
k--;
|
||
}
|
||
}
|
||
return 0;
|
||
}
|
||
|
||
public int getCount(int num, long n) {
|
||
int count = 0;
|
||
long min = num;
|
||
long max = num;
|
||
while (min <= n) {
|
||
count += Math.min(max, n) - min + 1;
|
||
min = min * 10;
|
||
max = max * 10 + 9;
|
||
}
|
||
return count;
|
||
}
|
||
|
||
}
|
||
//leetcode submit region end(Prohibit modification and deletion)
|
||
|
||
}
|