166. Fraction to Recurring Decimal
题目:
Given two integers representing the numerator and denominator of a fraction, return the fraction in string format.
If the fractional part is repeating, enclose the repeating part in parentheses.
For example,
- Given numerator = 1, denominator = 2, return "0.5".
- Given numerator = 2, denominator = 1, return "2".
- Given numerator = 2, denominator = 3, return "0.(6)".
链接: http://leetcode.com/problems/fraction-to-recurring-decimal/
题解:
这道题基本就是Divide Two Integers + 处理循环小数。一刷的时候果断放弃了,因为不想处理恶心的边界条件,不在一开始把int转为long的话后面会很难写。现在又看了一遍大家的答案,发现绝大多数是先把int转为long,我也就放心了。实在不行咱可以用Python写这个。
Java:
Time Complexity - O(n), Space Complexity - O(n)
public class Solution { public String fractionToDecimal(int numerator, int denominator) { if (numerator == 0) return "0"; if (denominator == 0) throw new ArithmeticException(); boolean sameSign = (numerator > 0) ^ (denominator < 0); long dividend = Math.abs((long)numerator); long divisor = Math.abs((long)denominator); String intPart = String.valueOf(dividend / divisor); intPart = sameSign ? intPart : "-" + intPart; long remainder = dividend % divisor; if (remainder == 0) return intPart; Map<Long, Integer> map = new HashMap<>(); StringBuilder decimalPart = new StringBuilder(); while (remainder != 0) { map.put(remainder, decimalPart.length()); remainder *= 10; decimalPart.append(remainder / divisor); remainder = remainder % divisor; if (map.containsKey(remainder)) { decimalPart.insert(map.get(remainder), "("); decimalPart.append(')'); break; } } return intPart + "." + decimalPart.toString(); } }
二刷:
一样的方法,一样的code,写起来还是很吃力,多多练习吧。
Java:
public class Solution { public String fractionToDecimal(int numerator, int denominator) { if (numerator == 0) return "0"; if (denominator == 0) throw new ArithmeticException("Divide by zero"); boolean sameSign = (numerator > 0) ^ (denominator < 0); long dividend = Math.abs((long)numerator); long divisor = Math.abs((long)denominator);
StringBuilder sb = new StringBuilder(); if (!sameSign) sb.append("-"); sb.append(dividend / divisor); dividend %= divisor; if (dividend == 0) return sb.toString(); sb.append("."); Map<Long, Integer> map = new HashMap<>(); while (dividend != 0) { map.put(dividend, sb.length()); dividend *= 10; sb.append(dividend / divisor); dividend %= divisor; if (map.containsKey(dividend)) { sb.insert(map.get(dividend), "("); sb.append(")"); return sb.toString(); } } return sb.toString(); } }
Test Cases:
- (1, 3)
- (1, 5)
- (1, 6)
- (1, 90)
- (1, 99)
- (22, 7)
- (-50, 8)
- (0, -5)
- (-1, -2147483648)
- (-2147483648, 1)
Reference:
https://leetcode.com/discuss/18731/accepted-cpp-solution-with-explainations
https://leetcode.com/discuss/18769/there-good-deal-with-extreme-edge-case-without-converting-long
https://leetcode.com/discuss/18989/online-judge-pass-java-version
https://leetcode.com/discuss/20515/my-java-solution
https://leetcode.com/discuss/22652/do-not-use-python-as-cpp-heres-a-short-version-python-code
https://leetcode.com/discuss/23079/my-clean-java-solution
https://leetcode.com/discuss/31521/short-java-solution
https://leetcode.com/discuss/42159/0ms-c-solution-with-detailed-explanations
https://leetcode.com/discuss/50512/accepted-clean-java-solution
http://gqqnbig.me/?p=160
http://blog.csdn.net/hanshileiai/article/details/8861376
http://segmentfault.com/q/1010000003958185
https://leetcode.com/discuss/8886/my-simple-solution
http://blog.csdn.net/ljiabin/article/details/42025037