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## Problem

Given a string expression representing an expression of fraction addition and subtraction, return the calculation result in string format.

The final result should be an irreducible fraction. If your final result is an integer, change it to the format of a fraction that has a denominator 1. So in this case, 2 should be converted to 2/1.

Example 1:

Input: expression = "-1/2+1/2"
Output: "0/1"


Example 2:

Input: expression = "-1/2+1/2+1/3"
Output: "1/3"


Example 3:

Input: expression = "1/3-1/2"
Output: "-1/6"


Constraints:

• The input string only contains ‘0’ to ‘9’, ‘/’, ‘+’ and ‘-‘. So does the output.
• Each fraction (input and output) has the 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.

## Code

365. Water and Jug Problem

Irreducible fraction 最简分数

Lookahead and lookbehind, collectively called “lookaround”

Look ahead positive (?=): find expression A where expression B follows:
A(?=B)

Look ahead negative (?!): find expression A where expression B does not follow:
A(?!B)

Look behind positive (?<=): find expression A where expression B precedes:
(?<=B)A

Look behind negative (?<!): find expression A where expression B does not precede:
(?<!B)A

"a;b;c;d".split("(?<=;)");
[a;, b;, c;, d]

"a;b;c;d".split("(?=;)");
[a, ;b, ;c, ;d]


640. Solve the Equation

class Solution {
public String fractionAddition(String expression) {
// | -> OR operator
String[] fracs = expression.split("/|(?=[+-])");

int i = 0;
int A = 0;
int B = 1;

while (i < fracs.length) {
int a = Integer.valueOf(fracs[i]);
int b = Integer.valueOf(fracs[i + 1]);

A = A * b + a * B;
B *= b;

int g = gcd(Math.abs(A), Math.abs(B));
A /= g;
B /= g;

i += 2;
}

return A + "/" + B;
}

// 两个正整数x和y(x>y)的最大公约数, 等于其中较小的数(y)和两数相除余数(x%y)的最大公约数
private int gcd(int x, int y) {
return (x % y == 0) ? y : gcd(y, x % y);
}
}