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:
- low ≤ high (注意二分循环退出的条件是小于等于)
- mid = low + (high-low)>>1 (防止溢出)
- 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)
参考代码
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
}
func abs(a int) int {
if a > 0 {
return a
}
return -a
}
标签:Integers,return,divisor,Divide,Two,abs,&&,dividend,quotient
From: https://blog.51cto.com/u_16110811/7201238