[leetcode]592. Fraction Addition and Subtraction

题目链接：https://leetcode.com/problems/fraction-addition-and-subtraction/#/description
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:

The input string only contains

'0'

to

'9'

,

'/'

,

'+'

and

'-'

. So does the output.
Each fraction (input and output) has format

±numerator/denominator

. If the first input fraction or the output is positive, then

'+'

will be omitted.
The input only contains valid irreducible fractions, where the
numerator and denominator of each fraction will always be in the range [1,10]. If the denominator is 1, it means this fraction is actually an integer in a fraction format defined above.
The number of given fractions will be in the range [1,10].
The numerator and denominator of the final result are guaranteed to be valid and in the range of 32-bit int.

思路：Keep the overall result in

A / B

, read the next fraction into

a / b

. Their sum is

(Ab + aB) / Bb

(but cancel their greatest common divisor).

方法一：

 class Solution {
public:
string fractionAddition(string expression) {
istringstream ss(expression);
int A = 0, B = 1, a, b;
char _;
while (ss >> a >> _ >> b) {
A = A * b + a * B;
B *= b;
int g = abs(GCD(A, B));
A /= g;
B /= g;
}
return to_string(A) + '/' + to_string(B);
}
private:
int GCD(int a, int b )
{
return (b == 0) ? a : GCD(b, a % b);
}

};
 

 方法二：

class Solution {
public:
string fractionAddition(string expression)
{
int n = 0, d = 1, p = 0, p1 = 0, p2 = 0; // n为上一个分子，d为上一个分母
if (expression[0] != '-') expression = "+" + expression;
while (p < expression.size())
{
for (p1 = p + 1; expression[p1] != '/'; ++p1);
for (p2 = p1 + 1; p2 < expression.size() && expression[p2] != '+' && expression[p2] != '-'; ++p2);
auto nn = stoi(expression.substr(p + 1, p1 - p - 1)), dd = stoi(expression.substr(p1 + 1, p2 - p1 - 1));// nn为当前分子，dd为当前分母
cout<<nn<<" "<<dd<<endl;
n=n*dd+(expression[p] == '-' ? -1 : 1)*nn*d;
d*=dd;
int g=abs(GCD(n,d));
n/=g;
d/=g;
p=p2;
}
return to_string(n) + "/" + to_string(d);

}
private:
int GCD(int a, int b )
{
return (b == 0) ? a : GCD(b, a % b);
}

};