python3 解法, 执行用时: 32 ms, 内存消耗: 14.9 MB, 提交时间: 2022-12-14 12:41:47

class Solution:
    def fractionAddition(self, expression: str) -> str:
        # 可以不单写成函数，放在同一个循环里更简洁
        def read_number(i: int):
            sign, numerator, denominator = -1 if expression[i] == '-' else 1, 0, 0
            if expression[i] in "+-":
                i += 1
            while i < len(expression) and expression[i] != '/':
                numerator = 10 * numerator + int(expression[i])
                i += 1
            i += 1
            while i < len(expression) and expression[i] not in "+-":
                denominator = 10 * denominator + int(expression[i])
                i += 1
            return sign * numerator, denominator if denominator else 1, i
        
        idx = numer = 0
        denom = 1
        while idx < len(expression):
            num, den, idx = read_number(idx)
            lm = lcm(denom, den)
            numer, denom = numer * lm // denom + num * lm // den, lm
        g = gcd(abs(numer), denom)
        return f"{numer // g}/{denom // g}"

javascript 解法, 执行用时: 64 ms, 内存消耗: 41.1 MB, 提交时间: 2022-12-14 12:41:21

/**
 * @param {string} expression
 * @return {string}
 */
var fractionAddition = function(expression) {
    let x = 0, y = 1; // 分子，分母
    let index = 0, n = expression.length;
    while (index < n) {
        // 读取分子
        let x1 = 0, sign = 1;
        if (expression[index] === '-' || expression[index] === '+') {
            sign = expression[index] === '-' ? -1 : 1;
            index++;
        }
        while (index < n && isDigit(expression[index])) {
            x1 = x1 * 10 + expression[index].charCodeAt() - '0'.charCodeAt();
            index++;
        }
        x1 = sign * x1;
        index++;

        // 读取分母
        let y1 = 0;
        while (index < n && isDigit(expression[index])) {
            y1 = y1 * 10 + expression[index].charCodeAt() - '0'.charCodeAt();
            index++;
        }

        x = x * y1 + x1 * y;
        y *= y1;
    }
    if (x === 0) {
        return "0/1";
    }
    const g = gcd(Math.abs(x), y); // 获取最大公约数
    return Math.floor(x / g) + "/" + Math.floor(y / g);
}

const gcd = (a, b) => {
    let remainder = a % b;
    while (remainder !== 0) {
        a = b;
        b = remainder;
        remainder = a % b;
    }
    return b;
};

const isDigit = (ch) => {
    return parseFloat(ch).toString() === "NaN" ? false : true;
}

golang 解法, 执行用时: 0 ms, 内存消耗: 1.9 MB, 提交时间: 2022-12-14 12:41:06

func fractionAddition(expression string) string {
    x, y := 0, 1 // 分子，分母
    for i, n := 0, len(expression); i < n; {
        // 读取分子
        x1, sign := 0, 1
        if expression[i] == '-' || expression[i] == '+' {
            if expression[i] == '-' {
                sign = -1
            }
            i++
        }
        for i < n && unicode.IsDigit(rune(expression[i])) {
            x1 = x1*10 + int(expression[i]-'0')
            i++
        }
        x1 = sign * x1
        i++

        // 读取分母
        y1 := 0
        for i < n && unicode.IsDigit(rune(expression[i])) {
            y1 = y1*10 + int(expression[i]-'0')
            i++
        }

        x = x*y1 + x1*y
        y *= y1
    }
    if x == 0 {
        return "0/1"
    }
    g := gcd(abs(x), y)
    return fmt.Sprintf("%d/%d", x/g, y/g)
}

func gcd(a, b int) int {
    for a != 0 {
        a, b = b%a, a
    }
    return b
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

java 解法, 执行用时: 4 ms, 内存消耗: 39.9 MB, 提交时间: 2022-12-14 12:40:50

class Solution {
    public String fractionAddition(String expression) {
        long x = 0, y = 1; // 分子，分母
        int index = 0, n = expression.length();
        while (index < n) {
            // 读取分子
            long x1 = 0, sign = 1;
            if (expression.charAt(index) == '-' || expression.charAt(index) == '+') {
                sign = expression.charAt(index) == '-' ? -1 : 1;
                index++;
            }
            while (index < n && Character.isDigit(expression.charAt(index))) {
                x1 = x1 * 10 + expression.charAt(index) - '0';
                index++;
            }
            x1 = sign * x1;
            index++;

            // 读取分母
            long y1 = 0;
            while (index < n && Character.isDigit(expression.charAt(index))) {
                y1 = y1 * 10 + expression.charAt(index) - '0';
                index++;
            }

            x = x * y1 + x1 * y;
            y *= y1;
        }
        if (x == 0) {
            return "0/1";
        }
        long g = gcd(Math.abs(x), y); // 获取最大公约数
        return Long.toString(x / g) + "/" + Long.toString(y / g);
    }

    public long gcd(long a, long b) {
        long remainder = a % b;
        while (remainder != 0) {
            a = b;
            b = remainder;
            remainder = a % b;
        }
        return b;
    }
}

python3 解法, 执行用时: 36 ms, 内存消耗: 14.9 MB, 提交时间: 2022-12-14 12:39:55

class Solution:
    def fractionAddition(self, expression: str) -> str:
        x, y = 0, 1  # 分子，分母
        i, n = 0, len(expression)
        while i < n:
            # 读取分子
            x1, sign = 0, 1
            if expression[i] == '-' or expression[i] == '+':
                if expression[i] == '-':
                    sign = -1
                i += 1
            while i < n and expression[i].isdigit():
                x1 = x1 * 10 + int(expression[i])
                i += 1
            x1 = sign * x1
            i += 1

            # 读取分母
            y1 = 0
            while i < n and expression[i].isdigit():
                y1 = y1 * 10 + int(expression[i])
                i += 1

            x = x * y1 + x1 * y
            y *= y1
        if x == 0:
            return "0/1"
        g = gcd(abs(x), y)
        return f"{x // g}/{y // g}"