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 in s are digits.

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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。

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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;
    }
}