添加内容

website/content/ChapterFour/0029.Divide-Two-Integers.md

@@ -0,0 +1,148 @@
+# [29. Divide Two Integers](https://leetcode.com/problems/divide-two-integers/)
+
+
+## 题目
+
+Given two integers `dividend` and `divisor`, divide two integers without using multiplication, division and mod operator.
+
+Return the quotient after dividing `dividend` by `divisor`.
+
+The integer division should truncate toward zero.
+
+**Example 1**:
+
+    Input: dividend = 10, divisor = 3
+    Output: 3
+
+**Example 2**:
+
+    Input: dividend = 7, divisor = -3
+    Output: -2
+
+**Note**:
+
+- Both dividend and divisor will be 32-bit signed integers.
+- The divisor will never be 0.
+- Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 2^31 − 1 when the division result overflows.
+
+
+## 题目大意
+
+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。
+
+说明:
+
+- 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 [−2^31,  2^31 − 1]。本题中，如果除法结果溢出，则返回 2^31 − 1。
+
+
+## 解题思路
+
+- 给出除数和被除数，要求计算除法运算以后的商。注意值的取值范围在 [−2^31, 2^31 − 1] 之中。超过范围的都按边界计算。
+- 这一题可以用二分搜索来做。要求除法运算之后的商，把商作为要搜索的目标。商的取值范围是 [0, dividend]，所以从 0 到被除数之间搜索。利用二分，找到(商 + 1 ) * 除数 > 被除数并且 商 * 除数 ≤ 被除数 或者 (商+1)* 除数 ≥ 被除数并且商 * 除数 < 被除数的时候，就算找到了商，其余情况继续二分即可。最后还要注意符号和题目规定的 Int32 取值范围。
+- 二分的写法常写错的 3 点：
+    1. low ≤ high (注意二分循环退出的条件是小于等于)
+    2. mid = low + (high-low)>>1 (防止溢出)
+    3. low = mid + 1 ; high = mid - 1 (注意更新 low 和 high 的值，如果更新不对就会死循环)
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"math"
+)
+
+// 解法一 递归版的二分搜索
+func divide(dividend int, divisor int) int {
+	sign, res := -1, 0
+	// low, high := 0, abs(dividend)
+	if dividend == 0 {
+		return 0
+	}
+	if divisor == 1 {
+		return dividend
+	}
+	if dividend == math.MinInt32 && divisor == -1 {
+		return math.MaxInt32
+	}
+	if dividend > 0 && divisor > 0 || dividend < 0 && divisor < 0 {
+		sign = 1
+	}
+	if dividend > math.MaxInt32 {
+		dividend = math.MaxInt32
+	}
+	// 如果把递归改成非递归，可以改成下面这段代码
+	// for low <= high {
+	// 	quotient := low + (high-low)>>1
+	// 	if ((quotient+1)*abs(divisor) > abs(dividend) && quotient*abs(divisor) <= abs(dividend)) || ((quotient+1)*abs(divisor) >= abs(dividend) && quotient*abs(divisor) < abs(dividend)) {
+	// 		if (quotient+1)*abs(divisor) == abs(dividend) {
+	// 			res = quotient + 1
+	// 			break
+	// 		}
+	// 		res = quotient
+	// 		break
+	// 	}
+	// 	if (quotient+1)*abs(divisor) > abs(dividend) && quotient*abs(divisor) > abs(dividend) {
+	// 		high = quotient - 1
+	// 	}
+	// 	if (quotient+1)*abs(divisor) < abs(dividend) && quotient*abs(divisor) < abs(dividend) {
+	// 		low = quotient + 1
+	// 	}
+	// }
+	res = binarySearchQuotient(0, abs(dividend), abs(divisor), abs(dividend))
+	if res > math.MaxInt32 {
+		return sign * math.MaxInt32
+	}
+	if res < math.MinInt32 {
+		return sign * math.MinInt32
+	}
+	return sign * res
+}
+
+func binarySearchQuotient(low, high, val, dividend int) int {
+	quotient := low + (high-low)>>1
+	if ((quotient+1)*val > dividend && quotient*val <= dividend) || ((quotient+1)*val >= dividend && quotient*val < dividend) {
+		if (quotient+1)*val == dividend {
+			return quotient + 1
+		}
+		return quotient
+	}
+	if (quotient+1)*val > dividend && quotient*val > dividend {
+		return binarySearchQuotient(low, quotient-1, val, dividend)
+	}
+	if (quotient+1)*val < dividend && quotient*val < dividend {
+		return binarySearchQuotient(quotient+1, high, val, dividend)
+	}
+	return 0
+}
+
+// 解法二 非递归版的二分搜索
+func divide1(divided int, divisor int) int {
+	if divided == math.MinInt32 && divisor == -1 {
+		return math.MaxInt32
+	}
+	result := 0
+	sign := -1
+	if divided > 0 && divisor > 0 || divided < 0 && divisor < 0 {
+		sign = 1
+	}
+	dvd, dvs := abs(divided), abs(divisor)
+	for dvd >= dvs {
+		temp := dvs
+		m := 1
+		for temp<<1 <= dvd {
+			temp <<= 1
+			m <<= 1
+		}
+		dvd -= temp
+		result += m
+	}
+	return sign * result
+}
+
+
+```