816. Ambiguous Coordinates**
https://leetcode.com/problems/ambiguous-coordinates/
题目描述
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 occurring 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: "(123)"
Output: ["(1, 23)", "(12, 3)", "(1.2, 3)", "(1, 2.3)"]
Example 2:
Input: "(00011)"
Output: ["(0.001, 1)", "(0, 0.011)"]
Explanation:
0.0, 00, 0001 or 00.01 are not allowed.
Example 3:
Input: "(0123)"
Output: ["(0, 123)", "(0, 12.3)", "(0, 1.23)", "(0.1, 23)", "(0.1, 2.3)", "(0.12, 3)"]
Example 4:
Input: "(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.
题目描述
这道题不算难, 但需要考虑到很多特殊情形. 我是这样思考的, 对于一个字符串 S, 首先可以将它分为两个整数部分, 这一步骤可以看做是在 S 中插入逗号. 然后对于这两个整数字符串, 可以检查它们各自可以形成多少个有效的数字(包括整数和浮点数), 这里牵涉两个难点:
对于第一点, 需要考虑的情况有:
123 # 正常
012 # 0 开头, 属于特殊情况
000 # 全零 (也属于 0 开头这种特殊情况)
120 # 0 结尾
对于 0 开头这个特殊情况, 可以使用:
if (std::to_string(std::stoi(s)) != s)
来鉴别. 而对于第二点, 需要考虑的情况为:
123 # 正常, 在数字间任意位置插入小数点都能形成有效数字
012 # 0 开头但不是 0 结尾, 属于特殊情况, 只能在第一个 0 后面插入一个小数点
010 # 0 开头并且以 0 结尾, 属于特殊情况, 不能形成有效数字
000 # 和上一种情况一样, 0 开头并且以 0 结尾
120 # 0 结尾, 只能形成整数的结果, 不能形成浮点数
处理方法如下:
class Solution {
private:
vector<string> extend(const string &s) {
vector<string> res;
// 处理以 0 开头的情况
if (std::to_string(std::stoi(s)) != s) {
if (s.back() != '0') {
res.push_back(s.substr(0, 1) + "." + s.substr(1, s.size() - 1));
}
return res;
}
res.push_back(s); // 保留正常的整数结果
if (s.back() == '0') return res; // 以 0 结尾, 直接返回上一步得到的整数结果
// 正常情况, 加浮点数
for (int i = 0; i < s.size() - 1; ++ i) {
string decimal_num = s.substr(0, i + 1) + "." + s.substr(i + 1, s.size() - (i + 1));
res.push_back(decimal_num);
}
return res;
}
public:
vector<string> ambiguousCoordinates(string S) {
vector<string> res;
S = S.substr(1, S.size() - 2);
for (int i = 0; i < S.size() - 1; ++ i) {
auto first = S.substr(0, i + 1);
auto second = S.substr(i + 1, S.size() - (i + 1));
auto A = extend(first), B = extend(second);
for (auto &a : A)
for (auto &b : B)
res.push_back("(" + a + ", " + b + ")");
}
return res;
}
};
