29. Divide Two Integers
给定两个整数,被除数 dividend 和除数 divisor。将两数相除,要求不使用乘法、除法和 mod 运算符。
返回被除数 dividend 除以除数 divisor 得到的商。
整数除法的结果应当截去(truncate)其小数部分,例如:truncate(8.345) = 8 以及 truncate(-2.7335) = -2
示例 1:
输入: dividend = 10, divisor = 3
输出: 3
解释: 10/3 = truncate(3.33333..) = truncate(3) = 3
示例 2:
输入: dividend = 7, divisor = -3
输出: -2
解释: 7/-3 = truncate(-2.33333..) = -2
提示:
被除数和除数均为 32 位有符号整数。 除数不为 0。 假设我们的环境只能存储 32 位有符号整数,其数值范围是 [−231, 231 − 1]。本题中,如果除法结果溢出,则返回 231 − 1。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/divide-two-integers/
Link:https://leetcode.com/problems/divide-two-integers/
暴力破解
题目限制,就只能用加减法,加上二进制操作了。暴力破解是每次减去divisor, 减了多少次, 结果就是多少。
难点在于,符号,边界条件判断,负数最大值转成正数就越界了。
不出意外的,暴力破解会超时
MAX_INT = 2147483647
MIN_INT = -2147483648
class Solution:
def divide(self, dividend: int, divisor: int) -> int:
if dividend == MIN_INT and divisor == -1:
return MAX_INT
res = 0
minus = (dividend > 0) ^ (divisor > 0)
dividend = abs(dividend)
divisor = abs(divisor)
while dividend >= divisor:
res += 1
dividend -= divisor
return -res if minus > 0 else res
二分法
每次减去一个divisor太慢了,我们可以尝试,减去(2, 4, 8, 16 …)倍的divisor
python代码如下:
MAX_INT = 2147483647
MIN_INT = -2147483648
class Solution:
def divide(self, dividend: int, divisor: int) -> int:
if dividend == MIN_INT and divisor == -1:
return MAX_INT
res = 0
minus = (dividend > 0) ^ (divisor > 0)
dividend = abs(dividend)
divisor = abs(divisor)
while dividend >= divisor:
tmp = divisor
scale = 1
while (dividend >= (tmp << 1)):
tmp = tmp << 1
scale = scale << 1
res += scale
dividend -= tmp
return -res if minus > 0 else res
C++代码如下:
class Solution {
public:
int divide(int dividend, int divisor) {
if (dividend == INT_MIN && divisor == -1) {
return INT_MAX;
}
bool minus = ((dividend > 0) ^ (divisor > 0));
long long number = labs(dividend);
long long div = labs(divisor);
long long res = 0;
while (number >= div) {
long long tmp = div;
long long mul = 1;
while(number >= (tmp << 1)) {
mul <<= 1;
tmp <<= 1;
}
res += mul;
number -= tmp;
}
return (int)(minus ? -res: res);
}
};
–End–