Algorithm
Problem Name: 29. Divide Two Integers
Problem Link: https://leetcode.com/problems/divide-two-integers/
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 - 1divisor != 0
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
private static int HALF_INT_MIN = -1073741824;
public int divide(int dividend, int divisor) {
if (dividend == Integer.MIN_VALUE && divisor == -1) {
return Integer.MAX_VALUE;
}
int negatives = 2;
if (dividend > 0) {
negatives--;
dividend = -dividend;
}
if (divisor > 0) {
negatives--;
divisor = -divisor;
}
int quotient = 0;
while (divisor >= dividend) {
int powerOfTwo = -1;
int value = divisor;
while (value >= HALF_INT_MIN && value + value >= dividend) {
value += value;
powerOfTwo += powerOfTwo;
}
quotient += powerOfTwo;
dividend -= value;
}
if (negatives != 1) {
return -quotient;
}
return quotient;
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const divide = function(dividend, divisor) {
if (!divisor || (dividend === Number.MIN_SAFE_INTEGER && divisor === -1)) {
return Number.MAX_SAFE_INTEGER;
}
const MAX_INT = Math.pow(2, 31) - 1;
if (dividend === -2147483648 && divisor === -1) return MAX_INT;
const sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1;
let dvd = Math.abs(dividend);
let dvs = Math.abs(divisor);
let res = 0;
while (dvd >= dvs) {
let tmp = dvs;
let multiple = 1;
while (dvd >= tmp << 1 && tmp << 1 > 0) {
tmp <<= 1;
multiple <<= 1;
}
dvd -= tmp;
res += multiple;
}
return sign === 1 ? res : -res;
};
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def divide(self, dividend, divisor):
"""
:type dividend: int
:type divisor: int
:rtype: int
"""
positive = (dividend < 0) is (divisor < 0)
dividend, divisor, res = abs(dividend), abs(divisor), 0
while dividend >= divisor:
temp, i = divisor, 1
while dividend >= temp:
dividend -= temp
res += i
i <<= 1
temp <<= 1
if not positive: res = -res
return min(max(-2 ** 31, res), 2 ** 31 - 1)
Input
Output
#4 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _029_DivideTwoIntegers
{
public int Divide(int dividend, int divisor)
{
if (divisor == 0) { throw new DivideByZeroException(); }
uint a = dividend > 0 ? (uint)dividend : (uint)-dividend;
uint b = divisor > 0 ? (uint)divisor : (uint)-divisor;
uint result = 0, c = 0;
var index = 0;
while (a >= b)
{
c = b;
for (index = 0; a >= c && c != 0; index++, c *= 2)
{
a -= c;
result += (uint)1 << index;
}
}
return (dividend ^ divisor) >> 31 == -1
? (int)-result
: result > int.MaxValue ? int.MaxValue : (int)result;
}
}
}
Input
Output