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【坚持每日一题9.23】运算

作者:互联网

请实现整数数字的乘法、减法和除法运算,运算结果均为整数数字,程序中只允许使用加法运算符和逻辑运算符,允许程序中出现正负常数,不允许使用位运算。

你的实现应该支持如下操作:

Operations() 构造函数
minus(a, b) 减法,返回a - b
multiply(a, b) 乘法,返回a * b
divide(a, b) 除法,返回a / b
示例:

Operations operations = new Operations();
operations.minus(1, 2); //返回-1
operations.multiply(3, 4); //返回12
operations.divide(5, -2); //返回-2
提示:

你可以假设函数输入一定是有效的,例如不会出现除法分母为0的情况
单个用例的函数调用次数不会超过1000次

java代码:

class Operations {

    // 用来获取-1
    int ne = Integer.MAX_VALUE + Integer.MAX_VALUE + 1;

    long[] neCache = new long[32];// 放置 -1,-2,-4,-8...
    long[] poCache = new long[32];// 放置 1,2,4,8...
    long[] cache = new long[32];// 存放乘数或除数的倍数,1*a,2*a,4*a,8*a...主要用于快速计算,不然容易超时
    long[] cache1 = new long[32];// 存放乘数或除数的倍数 负数-1*a,-2*a,-4*a,-8*a

    public Operations() {
        neCache[0] = ne;
        poCache[0] = 1;
        for (int i = 1; i < 32; ++i) {
            neCache[i] = neCache[i + ne] + neCache[i + ne];
            poCache[i] = poCache[i + ne] + poCache[i + ne];
        }
    }

    public int minus(int a, int b) {
        if (a == b) return 0;
        int index = 31;// 从最大值开始比较
        while (b != 0) {
            if (b > 0) {
                if (b >= poCache[index]) { // 如果b大于2的index次方,
                    b += neCache[index];// a与b同时减
                    a += neCache[index];
                } else {
                    index += ne;
                }
            } else { // b小于0时同理
                if (b <= neCache[index]) {
                    b += poCache[index];
                    a += poCache[index];
                } else {
                    index += ne;
                }
            }
        }
        return a;
    }

    public int multiply(int a, int b) {
        if (a == 0 || b == 0) return 0;
        if (a == 1) return b;
        if (b == 1) return a;
        if (a == ne) return minus(0, b);
        if (b == ne) return minus(0, a);
        int sign = (a > 0 && b > 0) || (a < 0 && b < 0) ? 1 : ne;
        // 把b变成正数
        if (b < 0) {
            b = minus(0, b);
        }

        cache[0] = a;
        for (int i = 1; i < 32; i++) {
            cache[i] = cache[i + ne] + cache[i + ne];
        }
        int index = 30; // 从31开始应该也是可以的
        int ret = 0;
        int retSign = a > 0 ? 1 : ne; // 记录返回值的符号
        while (b > 0) {
            if (b >= poCache[index]) {
                b += neCache[index];
                ret += cache[index];
                retSign = ret > 0 ? 1 : ne;// 记录返回值的符号
            } else {
                index += ne;
            }
        }
        // 根据初始值改变返回值的符号
        if ((sign < 0 && ret > 0) || (sign > 0 && ret < 0)) {
            ret = minus(0, ret);
        }
        // 结果溢出,返回值的符号会变成相反的
        if (retSign != (a > 0 ? 1 : ne)) {
            ret = minus(0, ret);
        }
        return ret;
    }

    public int divide(int a, int b) {
        if (a == 0) return 0;
        if (b == 1) return a;
        if (b == ne) return minus(0, a);
        int ret = 0;
        int sign = (a > 0 && b > 0) || (a < 0 && b < 0) ? 1 : ne;
        long nb = b;
        long pb = b;
        if (b < 0) {
            b = minus(0, b);
        } else {
            nb = minus(0, b);
        }
        if (a < 0) {
            a = minus(0, a);
        }
        cache[0] = b;
        cache1[0] = nb;
        int index = 1;
        for (; index < 32; ++index) {
            cache[index] = cache[index + ne] + cache[index + ne];
            cache1[index] = cache1[index + ne] + cache1[index + ne];
            if (cache1[index] >= a) {
                break; // 找到最大值就可以返回了,不用计算完
            }
        }
        if (index >= 32) index = 31;
        while (a >= b) {
            if (a >= cache[index]) {
                ret += poCache[index];// 注意这里是2的index次方的值
                a += cache1[index];
            } else {
                index += ne;
            }
        }
        if (sign < 0) {
            ret = minus(0, ret);
        }
        return ret;
    }
}


/**
 * Your Operations object will be instantiated and called as such:
 * Operations obj = new Operations();
 * int param_1 = obj.minus(a,b);
 * int param_2 = obj.multiply(a,b);
 * int param_3 = obj.divide(a,b);
 */

标签:9.23,index,运算,int,cache,ne,ret,一题,minus
来源: https://blog.csdn.net/kangbin825/article/details/120427609