移位操作搞定兩數之商

freephp發表於2024-05-05

五一漫長的假期，外面的世界是人山人海，反而在家刷題算得上一個好的休閒方式。剛好我開始寫這道題：

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`

看懂題目上說的，就是不能用乘法、除法以及取餘操作來算出兩個給定整數的商。這個時候我想到利用移位操作來實現。
雖然工作多年，但是真正在實際專案中用到移位操作的時候是很少的。
邏輯移位：

-   邏輯移位將位向左或向右移動，並在空位填充零。
-   在左邏輯移位（<<）中，位向左移動，從右側填充零。
-   在右邏輯移位（>>）中，位向右移動，從左側填充零。

簡單來說，如果1<<2, 就是1乘以2的2次方，以此類推。
所以我的解法就很明確了, 處理好被除數和除數的符號，然後再透過迴圈裡面使用移位操作計算出商的：


func divide(dividend int, divisor int) int {
	quotient := 0
	maxInt := math.MaxInt32
	minInt := math.MinInt32
	if divisor == 0 || (dividend == minInt && divisor == -1) {
		return maxInt
	}

	// determine the sign of the quotient
	sign := -1
	if (dividend > 0 && divisor > 0) || (dividend < 0 && divisor < 0) {
		sign = 1
	}

	// Convert the dividend and divisor to be positive number
	positiveDividend, positiveDivisor := int(math.Abs(float64(dividend))), int(math.Abs(float64(divisor)))

	for positiveDividend >= positiveDivisor {
		shift := 0
		for positiveDividend >= (positiveDivisor << shift) {
			shift += 1
		}

		quotient += (1<< (shift - 1))
		positiveDividend -= (positiveDivisor << (shift - 1))
	}
	
	return int(math.Min(float64(maxInt), math.Max(float64(sign) * float64(quotient), float64(minInt))))
}

總結

移位操作雖然好，但也不是唯一解，回憶一下小時候還沒學乘法的時候，我們也可以用加法去模擬乘法，所以利用累加來模擬出兩數之商更直觀。
經過我的嘗試，我發現完全減法會導致超時問題，不得不結合shifting操作，程式碼如下所示：


func divide(dividend int, divisor int) int {
	quotient := 0
	maxInt := math.MaxInt32
	minInt := math.MinInt32
	if divisor == 0 || (dividend == minInt && divisor == -1) {
		return maxInt
	}
	// determine the sign of the quotient
	sign := -1
	if (dividend ^ divisor) >= 0 {
		sign = 1
	}

	// Convert the dividend and divisor to be positive number
	positiveDividend, positiveDivisor := int(math.Abs(float64(dividend))), int(math.Abs(float64(divisor)))

	// Every time postiveDividend subtract with the value of the divisor
	for positiveDividend >= positiveDivisor {
		// Initialize variables for binary search
        tempDivisor := positiveDivisor
        multiple := 1
        
        // Perform binary search-like division
        for positiveDividend >= (tempDivisor << 1) {
            tempDivisor <<= 1
            multiple <<= 1
        }
        
        // Subtract the multiple of divisor from dividend
        positiveDividend -= tempDivisor
        // Add the multiple to quotient
        quotient += multiple
	}
	return int(math.Min(float64(maxInt), math.Max(float64(sign) * float64(quotient), float64(minInt))))
 }

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