LeetCode 题解工作台
千位分隔数
给你一个整数 n ，请你每隔三位添加点（即 "." 符号）作为千位分隔符，并将结果以字符串格式返回。 示例 1： 输入： n = 987 输出： "987" 示例 2： 输入： n = 1234 输出： "1.234" 示例 3： 输入： n = 123456789 输出： "123.456.789…
1
题型
4
代码语言
3
相关题
当前训练重点
简单 · String-driven solution strategy
答案摘要
直接按照题目要求模拟即可。 时间复杂度 $O(\log n)$，忽略答案的空间消耗，空间复杂度 。
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题目描述
给你一个整数 n，请你每隔三位添加点（即 "." 符号）作为千位分隔符，并将结果以字符串格式返回。
示例 1：
输入：n = 987 输出："987"
示例 2：
输入：n = 1234 输出："1.234"
示例 3：
输入：n = 123456789 输出："123.456.789"
示例 4：
输入：n = 0 输出："0"
提示：
0 <= n < 2^31
解题思路
方法一：模拟
直接按照题目要求模拟即可。
时间复杂度 ，忽略答案的空间消耗，空间复杂度 。
class Solution:
def thousandSeparator(self, n: int) -> str:
cnt = 0
ans = []
while 1:
n, v = divmod(n, 10)
ans.append(str(v))
cnt += 1
if n == 0:
break
if cnt == 3:
ans.append('.')
cnt = 0
return ''.join(ans[::-1])
复杂度分析
| 指标 | 值 |
|---|---|
| 时间 | complexity is O(log n) because the number of digits in n is proportional to log(n). Space complexity is also O(log n) to store the intermediate string and reconstructed output with separators. |
| 空间 | Depends on the final approach |
面试官常问的追问
外企场景- question_mark
Pay attention to scanning direction; starting from the wrong end can misplace separators.
- question_mark
Watch for edge cases where n < 1000, as no separators should be added.
- question_mark
Confirm that the solution does not modify the numeric value, only its string representation.
常见陷阱
外企场景- error
Inserting separators from left to right without counting groups can create extra dots.
- error
Failing to handle numbers less than 1000 may produce incorrect output.
- error
Neglecting to reverse the string after inserting separators from the end leads to misaligned formatting.
进阶变体
外企场景- arrow_right_alt
Use commas ',' instead of dots '.' for different locale formatting.
- arrow_right_alt
Apply the separator pattern to floating-point numbers by splitting integer and fractional parts.
- arrow_right_alt
Implement a recursive solution that formats the number by repeatedly processing blocks of three digits.