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149 lines
4.5 KiB
Markdown
Executable File
149 lines
4.5 KiB
Markdown
Executable File
# [29. Divide Two Integers](https://leetcode.com/problems/divide-two-integers/)
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## 题目
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Given two integers `dividend` and `divisor`, divide two integers without using multiplication, division and mod operator.
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Return the quotient after dividing `dividend` by `divisor`.
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The integer division should truncate toward zero.
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**Example 1**:
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Input: dividend = 10, divisor = 3
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Output: 3
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**Example 2**:
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Input: dividend = 7, divisor = -3
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Output: -2
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**Note**:
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- Both dividend and divisor will be 32-bit signed integers.
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- The divisor will never be 0.
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- 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.
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## 题目大意
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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。
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说明:
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- 被除数和除数均为 32 位有符号整数。
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- 除数不为 0。
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- 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。
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## 解题思路
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- 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。
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- 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。
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- 二分的写法常写错的 3 点：
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1. low ≤ high (注意二分循环退出的条件是小于等于)
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2. mid = low + (high-low)>>1 (防止溢出)
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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)
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## 代码
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```go
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package leetcode
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import (
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"math"
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)
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// 解法一 递归版的二分搜索
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func divide(dividend int, divisor int) int {
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sign, res := -1, 0
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// low, high := 0, abs(dividend)
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if dividend == 0 {
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return 0
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}
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if divisor == 1 {
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return dividend
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}
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if dividend == math.MinInt32 && divisor == -1 {
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return math.MaxInt32
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}
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if dividend > 0 && divisor > 0 || dividend < 0 && divisor < 0 {
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sign = 1
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}
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if dividend > math.MaxInt32 {
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dividend = math.MaxInt32
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}
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// 如果把递归改成非递归，可以改成下面这段代码
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// for low <= high {
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// quotient := low + (high-low)>>1
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// if ((quotient+1)*abs(divisor) > abs(dividend) && quotient*abs(divisor) <= abs(dividend)) || ((quotient+1)*abs(divisor) >= abs(dividend) && quotient*abs(divisor) < abs(dividend)) {
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// if (quotient+1)*abs(divisor) == abs(dividend) {
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// res = quotient + 1
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// break
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// }
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// res = quotient
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// break
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// }
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// if (quotient+1)*abs(divisor) > abs(dividend) && quotient*abs(divisor) > abs(dividend) {
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// high = quotient - 1
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// }
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// if (quotient+1)*abs(divisor) < abs(dividend) && quotient*abs(divisor) < abs(dividend) {
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// low = quotient + 1
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// }
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// }
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res = binarySearchQuotient(0, abs(dividend), abs(divisor), abs(dividend))
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if res > math.MaxInt32 {
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return sign * math.MaxInt32
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}
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if res < math.MinInt32 {
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return sign * math.MinInt32
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}
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return sign * res
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}
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func binarySearchQuotient(low, high, val, dividend int) int {
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quotient := low + (high-low)>>1
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if ((quotient+1)*val > dividend && quotient*val <= dividend) || ((quotient+1)*val >= dividend && quotient*val < dividend) {
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if (quotient+1)*val == dividend {
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return quotient + 1
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}
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return quotient
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}
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if (quotient+1)*val > dividend && quotient*val > dividend {
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return binarySearchQuotient(low, quotient-1, val, dividend)
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}
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if (quotient+1)*val < dividend && quotient*val < dividend {
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return binarySearchQuotient(quotient+1, high, val, dividend)
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}
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return 0
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}
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// 解法二 非递归版的二分搜索
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func divide1(divided int, divisor int) int {
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if divided == math.MinInt32 && divisor == -1 {
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return math.MaxInt32
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}
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result := 0
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sign := -1
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if divided > 0 && divisor > 0 || divided < 0 && divisor < 0 {
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sign = 1
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}
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dvd, dvs := abs(divided), abs(divisor)
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for dvd >= dvs {
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temp := dvs
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m := 1
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for temp<<1 <= dvd {
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temp <<= 1
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m <<= 1
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}
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dvd -= temp
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result += m
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}
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return sign * result
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}
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```
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