标签：vector actual char fine and present back 0.00 actually
Given a string representing an expression of fraction addition and subtraction, you need to return the calculation result in string format. The final result should be irreducible fraction. If your final result is an integer, say 2, you need to change it to the format of fraction that has denominator 1. So in this case, 2 should be converted to 2/1.
Example 1:
Input:"-1/2+1/2" Output: "0/1"
Example 2:
Input:"-1/2+1/2+1/3" Output: "1/3"
Example 3:
Input:"1/3-1/2" Output: "-1/6"
Example 4:
Input:"5/3+1/3" Output: "2/1"
Note:
‘0‘ to ‘9‘, ‘/‘, ‘+‘ and ‘-‘. So does the output.±numerator/denominator. If the first input fraction or the output is positive, then ‘+‘ will be omitted.
class Solution { public: long long gcd(long long a, long long b) { if(a < b) return gcd(b,a); if(b == 0) return a; return gcd(b, a%b); } string fractionAddition(string expression) { vector<int> numerator; vector<int> denominator; int j = 0; bool positive = false; if(expression[0] >= ‘0‘ && expression[0] <= ‘9‘) positive = true; else { positive = false; j = 1; } string tmp; for(size_t i=j+1; i<expression.size(); i++) { if(expression[i] == ‘+‘ || expression[i] == ‘-‘) { tmp = expression.substr(j, i-j); size_t k; for(k = 0; k < tmp.size(); k++) { if(tmp[k] == ‘/‘) break; } numerator.push_back(stoi(tmp.substr(0,k))); if(!positive) numerator[numerator.size()-1] *= -1; positive = expression[i] == ‘+‘ ? true : false; denominator.push_back(stoi(tmp.substr(k+1))); j = i+1; } } tmp = expression.substr(j,expression.size()-j); size_t k; for(k=0; k<tmp.size(); k++) if(tmp[k] == ‘/‘) break; numerator.push_back(stoi(tmp.substr(0,k))); if(!positive) numerator[numerator.size()-1] *= -1; denominator.push_back(stoi(tmp.substr(k+1))); // test long long ret_numer = numerator[0]; long long ret_denomi = denominator[0]; long long r = 0; for(size_t i= 1; i<denominator.size(); i++) { long long tmp_deno = ret_denomi; ret_denomi *= denominator[i]; ret_numer = ret_numer * denominator[i] + tmp_deno * numerator[i]; } if(ret_numer > 0) { r = gcd(ret_numer, ret_denomi); ret_numer /= r; ret_denomi /= r; } else if (ret_numer < 0) { r = gcd(-ret_numer, ret_denomi); ret_numer /= r; ret_denomi /= r; } else if(ret_numer == 0) { ret_denomi = 1; } string str_numer = ret_numer < 0 ? ("-" + to_string(-ret_numer)) : to_string(ret_numer); string str_deno = to_string(ret_denomi); return str_numer + "/" + str_deno; } };
更好的一个思路
class Solution { public: string fractionAddition(string expression) { istringstream iss(expression); int num = 0, den = 0, NUM = 0, DEN = 1; char c; while (iss >> num >> c >> den) { NUM = NUM * den + num * DEN; DEN *= den; int g = abs(gcd(NUM, DEN)); NUM /= g; DEN /= g; } return to_string(NUM) + "/" + to_string(DEN); } int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); } };
LC 592. Fraction Addition and Subtraction
标签：vector actual char fine and present back 0.00 actually
原文地址：https://www.cnblogs.com/ethanhong/p/10354343.html
