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.
class Solution {
public String fractionAddition(String expression) {
List<Character> sign=new ArrayList<>();
for(int i=1;i<expression.length();i++){
if(expression.charAt(i)=='+'||expression.charAt(i)=='-'){
sign.add(expression.charAt(i));
}
}
List<Integer> up=new ArrayList<>();
List<Integer> down=new ArrayList<>();
for(String s1:expression.split("\\+")){
for(String s2:s1.split("-")){
if(s2.length()>0){
String []fraction=s2.split("/");
up.add(Integer.parseInt(fraction[0]));
down.add(Integer.parseInt(fraction[1]));
}
}
}
if(expression.charAt(0)=='-'){
up.set(0,-up.get(0));
}
int lcm=1;
for(int m:down){
lcm=lcm_(lcm,m);
}
int ans=lcm/down.get(0)*up.get(0);
for(int i=1;i<up.size();i++){
if(sign.get(i-1)=='+'){
ans+=lcm/down.get(i)*up.get(i);
}else{
ans-=lcm/down.get(i)*up.get(i);
}
}
int g=gcd(Math.abs(ans),Math.abs(lcm));
return (ans/g)+"/"+(lcm/g);
}
public int lcm_(int a,int b){
return a*b/gcd(a,b);
}
public int gcd(int a,int b){
while(b!=0){
int t=b;
b=a%b;
a=t;
}
return a;
}
}

本文介绍了一种算法,用于解析并计算分数加减的字符串表达式,返回最简分数形式的结果。通过将输入字符串分解为分子、分母和符号,使用最小公倍数和最大公约数的概念进行计算。
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