0816. Ambiguous Coordinates (M)
Ambiguous Coordinates (M)
题目
We had some 2-dimensional coordinates, like "(1, 3)"
or "(2, 0.5)"
. Then, we removed all commas, decimal points, and spaces, and ended up with the string s
. Return a list of strings representing all possibilities for what our original coordinates could have been.
Our original representation never had extraneous zeroes, so we never started with numbers like "00", "0.0", "0.00", "1.0", "001", "00.01", or any other number that can be represented with less digits. Also, a decimal point within a number never occurs without at least one digit occuring before it, so we never started with numbers like ".1".
The final answer list can be returned in any order. Also note that all coordinates in the final answer have exactly one space between them (occurring after the comma.)
Example 1:
Input: s = "(123)"
Output: ["(1, 23)", "(12, 3)", "(1.2, 3)", "(1, 2.3)"]
Example 2:
Input: s = "(00011)"
Output: ["(0.001, 1)", "(0, 0.011)"]
Explanation:
0.0, 00, 0001 or 00.01 are not allowed.
Example 3:
Input: s = "(0123)"
Output: ["(0, 123)", "(0, 12.3)", "(0, 1.23)", "(0.1, 23)", "(0.1, 2.3)", "(0.12, 3)"]
Example 4:
Input: s = "(100)"
Output: [(10, 0)]
Explanation:
1.0 is not allowed.
Note:
4 <= s.length <= 12
.s[0]
= "(",s[s.length - 1]
= ")", and the other elements ins
are digits.
题意
将一个数字字符串拆分为两个整数/小数,要求不能有多余的0。
思路
先将字符串拆分为左右两个部分,再在各个部分中插入小数点,判断是否存在多余的0,最后统计所有的结果。
代码实现
Java
class Solution {
public List<String> ambiguousCoordinates(String s) {
List<String> list = new ArrayList<>();
for (int i = 2; i < s.length() - 1; i++) {
for (String left : generate(s.substring(1, i))) {
for (String right : generate(s.substring(i, s.length() - 1))) {
list.add("(" + left + ", " + right + ")");
}
}
}
return list;
}
private List<String> generate(String s) {
List<String> list = new ArrayList<>();
for (int i = 1; i <= s.length(); i++) {
String left = s.substring(0, i);
String right = s.substring(i);
if ((left.equals("0") || !left.startsWith("0")) && !right.endsWith("0")) {
list.add(left + (i < s.length() ? "." : "") + right);
}
}
return list;
}
}
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 智能桌面机器人:用.NET IoT库控制舵机并多方法播放表情
· Linux glibc自带哈希表的用例及性能测试
· 深入理解 Mybatis 分库分表执行原理
· 如何打造一个高并发系统?
· .NET Core GC压缩(compact_phase)底层原理浅谈
· 新年开篇:在本地部署DeepSeek大模型实现联网增强的AI应用
· DeepSeek火爆全网,官网宕机?本地部署一个随便玩「LLM探索」
· Janus Pro:DeepSeek 开源革新,多模态 AI 的未来
· 上周热点回顾(1.20-1.26)
· 【译】.NET 升级助手现在支持升级到集中式包管理