Fraction Addition and Subtraction

问题：

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.

解决：

① 实现分数的加减。直接计算：A / B + a / b = (Ab + aB) / Bb

class Solution{ //105ms
public String fractionAddition(String expression) {
Scanner sc = new Scanner(expression).useDelimiter("/|(?=[-+])");
int A = 0;
int B = 1;
while(sc.hasNext()){
int a = sc.nextInt();
int b = sc.nextInt();
A = A * b + a * B;
B *= b;
int g = gcd(A,B);
A /= g;
B /= g;
}
return A + "/" + B;
}
public int gcd(int a,int b){
return a != 0 ? gcd(b % a,a) : Math.abs(b);
}
}

② 直接遍历计算。

class Solution{//11ms
public String fractionAddition(String expression) {
int i = 0;
int A = 0;
int B = 1;
while(i < expression.length()){
int sign = 1;
if (expression.charAt(i) == '-'){
sign = -1;
i ++;
}else if (expression.charAt(i) == '+'){
sign = 1;
i ++;
}
int a = 0;
while(Character.isDigit(expression.charAt(i))){
a = a * 10 + expression.charAt(i) - '0';
i ++;
}
a *= sign;
i ++;//跳过操作符
int b = 0;
while(i < expression.length() && Character.isDigit(expression.charAt(i))){
b = b * 10 + expression.charAt(i) - '0';
i ++;
}
int m = lcm(B,b);
A = A * m / B;
a = a * m / b;
A += a;
B = m;
}
int g = gcd(Math.abs(A),Math.abs(B));
A /= g;
B /= g;
return A + "/" + B;
}
public int lcm(int a,int b){
return a * b / gcd(a,b);
}
public int gcd(int a,int b){
return a != 0 ? gcd(b % a,a) : Math.abs(b);
}
}

③ 利用jdk1.8的新特性

class Solution {//113ms
public String fractionAddition(String expression) {//"(?=[-+])"表示匹配-或者+中的任意一个前面的字符。
String[] fractions = expression.split("(?=[-+])");
String res = "0/1";
for (String frac : fractions){
res = add(res,frac);
}
return res;
}
public String add(String frac1,String frac2){
int[] f1 = Stream.of(frac1.split("/")).mapToInt(Integer :: parseInt).toArray();
int[] f2 = Stream.of(frac2.split("/")).mapToInt(Integer :: parseInt).toArray();
int num = f1[0] * f2[1] + f1[1] * f2[0];
int den = f1[1] * f2[1];
String sign = "";
if (num < 0){
sign = "-";
num *= -1;
}
return sign + num / gcd(num,den) + "/" + den / gcd(num,den);
}
public int gcd(int x,int y){
return x == 0 || y == 0 ? x + y : gcd(y,x % y);
}
}

【注】