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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 stringS. 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: "(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 inSare digits.
Approach #1: Brute Force. [Java] [Memory Limit Exceeded]
class Solution { public List ambiguousCoordinates(String S) { List ans = new ArrayList<>(); StringBuilder sb = new StringBuilder(S); for (int i = 1; i < S.length(); ++i) { StringBuilder perfix = new StringBuilder(sb.substring(0, i)); StringBuilder suffix = new StringBuilder(sb.substring(i)); if (valid(perfix) && valid(suffix)) { List l1 = split(perfix); List l2 = split(suffix); for (int j = 0; j < l1.size(); ++j) { for (int k = 0; k < l2.size(); ++k) { String temp = "(" + l1.get(j) + ", " + l2.get(k) + ")"; ans.add(temp); } } } } return ans; } public List split(StringBuilder sb) { List ret = new ArrayList<>(); if (sb.length() == 1) { ret.add(sb.toString()); return ret; } else if (sb.charAt(0) == '0' && sb.charAt(1) == '0') { sb.insert(1, '.'); ret.add(sb.toString()); return ret; } else if (sb.charAt(sb.length() - 1) == '0' && sb.charAt(sb.length() - 2) == '0') { sb.insert(sb.length() - 1, '.'); ret.add(sb.toString()); return ret; } else { for (int i = 1; i < sb.length() - 1; ++i) { StringBuilder temp = sb; sb.insert(i, '.'); ret.add(sb.toString()); } } return ret; } public boolean valid(StringBuilder sb) { if (sb.length() == 1) return true; if (sb.length() > 4 && sb.charAt(0) == '0' && sb.charAt(1) == 0 && sb.charAt(sb.length() - 2) == '0' && sb.charAt(sb.length() - 1) == '0') return false; for (int i = 0; i < sb.length(); ++i) if (sb.charAt(i) != '0') return true; return false; }} ������
Approach #2: String. [Java]
class Solution { public List ambiguousCoordinates(String S) { List ans = new ArrayList<>(); int n = S.length(); for (int i = 1; i < n - 1; ++i) { List A = f(S.substring(1, i)), B = f(S.substring(i, n-1)); for (String a : A) for (String b : B) ans.add("(" + a + ", " + b + ")"); } return ans; } public List f(String s) { int n = s.length(); List ret = new ArrayList<>(); if (n == 0 || n > 1 && s.charAt(0) == '0' && s.charAt(n-1) == '0') return ret; if (n > 1 && s.charAt(0) == '0') { ret.add("0." + s.substring(1)); return ret; } ret.add(s); if (n == 1 || s.charAt(n-1) == '0') return ret; for (int i = 1; i < n; ++i) { ret.add(s.substring(0, i) + "." + s.substring(i, n)); } return ret; }}
Analysis:
if S == "" : return []
if S == "0" : return [S]
if S == "0XXXX0" : return []
if S == "0XXXX" : return ["0.XXXX"]
if S == "XXXX0" : return [S]
return [S, "X.XXX", "XX.XX", "XXX.X" ...]
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Reference:
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