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29. Divide Two Integers

中文文档

Description

Given two integers dividend and divisor, divide two integers without using multiplication, division, and mod operator.

The integer division should truncate toward zero, which means losing its fractional part. For example, 8.345 would be truncated to 8, and -2.7335 would be truncated to -2.

Return the quotient after dividing dividend by divisor.

Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For this problem, if the quotient is strictly greater than 231 - 1, then return 231 - 1, and if the quotient is strictly less than -231, then return -231.

 

Example 1:

Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = 3.33333.. which is truncated to 3.

Example 2:

Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = -2.33333.. which is truncated to -2.

 

Constraints:

  • -231 <= dividend, divisor <= 231 - 1
  • divisor != 0

Solutions

Approach 1: Quick Power

Time complexity $O(\log a \times \log b)$, Space complexity $O(1)$.

Python3

class Solution:
    def divide(self, a: int, b: int) -> int:
        INT_MAX = (1 << 31) - 1
        INT_MIN = -(1 << 31)
        sign = -1 if a * b < 0 else 1
        a = abs(a)
        b = abs(b)
        tot = 0
        while a >= b:
            cnt = 0
            while a >= (b << (cnt + 1)):
                cnt += 1
            tot += 1 << cnt
            a -= b << cnt
        return sign * tot if INT_MIN <= sign * tot <= INT_MAX else INT_MAX

Java

class Solution {
    public int divide(int a, int b) {
        int sign = 1;
        if ((a < 0) != (b < 0)) {
            sign = -1;
        }
        long x = Math.abs((long) a);
        long y = Math.abs((long) b);
        long tot = 0;
        while (x >= y) {
            int cnt = 0;
            while (x >= (y << (cnt + 1))) {
                cnt++;
            }
            tot += 1L << cnt;
            x -= y << cnt;
        }
        long ans = sign * tot;
        if (ans >= Integer.MIN_VALUE && ans <= Integer.MAX_VALUE) {
            return (int) ans;
        }
        return Integer.MAX_VALUE;
    }
}

Go

func divide(a int, b int) int {
	sign, ans, INT32_MAX, INT32_MIN, LIMIT := false, 0, 1<<31-1, -1<<31, -1<<31/2
	if (a > 0 && b < 0) || (a < 0 && b > 0) {
		sign = true
	}
	a, b = convert(a), convert(b)
	for a <= b {
		cnt := 0
		// (b<<cnt) >= LIMIT 是为了避免 b<<(cnt+1) 发生溢出
		for (b<<cnt) >= LIMIT && a <= (b<<(cnt+1)) {
			cnt++
		}
		ans = ans + -1<<cnt
		a = a - b<<cnt
	}
	if sign {
		return ans
	}
	if ans == INT32_MIN {
		return INT32_MAX
	}
	return -ans
}

func convert(v int) int {
	if v > 0 {
		return -v
	}
	return v
}

C++

class Solution {
public:
    int divide(int a, int b) {
        int sign = 1;
        if (a < 0 ^ b < 0) {
            sign = -1;
        }

        auto x = abs(static_cast<long long>(a));
        auto y = abs(static_cast<long long>(b));
        auto tot = 0ll;
        while (x >= y) {
            int cnt = 0;
            while (x >= (y << (cnt + 1))) {
                ++cnt;
            }
            tot += 1ll << cnt;
            x -= y << cnt;
        }

        auto ans = sign * tot;
        if (ans >= INT32_MIN && ans <= INT32_MAX) {
            return static_cast<int>(ans);
        }
        return INT32_MAX;
    }
};

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