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： 

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
     
       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
     
       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. (1, 3)
2. (1, 5)
3. (1, 6)
4. (1, 90)
5. (1, 99)
6. (22, 7)
7. (-50, 8)
8. (0, -5)
9. (-1, -2147483648)
10. (-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