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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.
class Solution {
public:
int findKthNumber(int n, int k) {
int cur = 1;
--k;
while (k > 0) {
long long step = 0, first = cur, last = cur + 1;
while (first <= n) {
step += min((long long)n + 1, last) - first;
first *= 10;
last *= 10;
}
if (step <= k) {
++cur;
k -= step;
} else {
cur *= 10;
--k;
}
}
return cur;
}
};
Given integers
n
andk
, find the lexicographically k-th smallest integer in the range from1
ton
.Note: 1 ≤ k ≤ n ≤ 109.
Example:
这道题是之前那道Lexicographical Numbers的延伸,之前让按字典顺序打印数组,而这道题让我们快速定位某一个位置,那么我们就不能像之前那道题一样,一个一个的遍历,这样无法通过OJ,这也是这道题被定为Hard的原因。那么我们得找出能够快速定位的方法,我们如果仔细观察字典顺序的数组,我们可以发现,其实这是个十叉树Denary Tree,就是每个节点的子节点可以有十个,比如数字1的子节点就是10到19,数字10的子节点可以是100到109,但是由于n大小的限制,构成的并不是一个满十叉树。我们分析题目中给的例子可以知道,数字1的子节点有4个(10,11,12,13),而后面的数字2到9都没有子节点,那么这道题实际上就变成了一个先序遍历十叉树的问题,那么难点就变成了如何计算出每个节点的子节点的个数,我们不停的用k减去子节点的个数,当k减到0的时候,当前位置的数字即为所求。现在我们来看如何求子节点个数,比如数字1和数字2,我们要求按字典遍历顺序从1到2需要经过多少个数字,首先把1本身这一个数字加到step中,然后我们把范围扩大十倍,范围变成10到20之前,但是由于我们要考虑n的大小,由于n为13,所以只有4个子节点,这样我们就知道从数字1遍历到数字2需要经过5个数字,然后我们看step是否小于等于k,如果是,我们cur自增1,k减去step;如果不是,说明要求的数字在子节点中,我们此时cur乘以10,k自减1,以此类推,直到k为0推出循环,此时cur即为所求:
类似题目:
Lexicographical Numbers
参考资料:
https://discuss.leetcode.com/topic/64624/concise-easy-to-understand-java-5ms-solution-with-explaination/2
https://discuss.leetcode.com/topic/64462/c-python-0ms-o-log-n-2-time-o-1-space-super-easy-solution-with-detailed-explanations
LeetCode All in One 题目讲解汇总(持续更新中...)
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