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:
解法
方法一
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20 | 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
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31 | 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;
}
}
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32 | 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;
}
};
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25 | 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
}
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