标签：== put inpu map 溢出 ret string find eating
算法描述：
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.
Example 1:
Input: numerator = 1, denominator = 2 Output: "0.5"
Example 2:
Input: numerator = 2, denominator = 1 Output: "2"
Example 3:
Input: numerator = 2, denominator = 3 Output: "0.(6)"
解题思路：模拟题。需要注意的点：1 正负号，2 溢出 ，3 小数点 ，4利用unordered_map判断重复
string fractionToDecimal(int numerator, int denominator) { if(numerator==0) return "0"; string res = ""; if((numerator <0 && denominator > 0 ) || (numerator >0 && denominator <0)) res += "-"; long long num1 = llabs(numerator); long long num2 = llabs(denominator); res += to_string(num1/num2); long long mo = num1%num2; if(mo == 0) return res; res += "."; unordered_map<int,int> unmap; while(mo > 0){ if(unmap.find(mo)!=unmap.end()){ res.insert(unmap[mo],1,‘(‘); res += ")"; break; } unmap[mo]=res.size(); res += to_string(mo*10 / num2); mo = mo*10 % num2; } return res; }
LeetCode-166- Fraction to Recurring Decimal
标签：== put inpu map 溢出 ret string find eating
原文地址：https://www.cnblogs.com/nobodywang/p/10400255.html
