Topic
- Math
Description
https://leetcode.com/problems/divide-two-integers/
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, which means losing its fractional part. For example, truncate(8.345) = 8 and truncate(-2.7335) = -2.
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, assume that your function returns 2³¹ − 1 when the division result overflows.
Example 1:
Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = truncate(3.33333..) = 3.
Example 2:
Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = truncate(-2.33333..) = -2.
Example 3:
Input: dividend = 0, divisor = 1
Output: 0
Example 4:
Input: dividend = 1, divisor = 1
Output: 1
Constraints:
- -2³¹ <= dividend, divisor <= 2³¹ - 1
- divisor != 0
Analysis
方法一：我寫的。
一開始，打算將被除數，除數絕對值後（先得出結果正負號），然後二分查找一個候選商，然後除數累加商值數，累加值與被除數相比，從而得出最佳候選商。但這方法在溢出情況處理很棘手，特別是被除數是Integer.MIN_VALUE情況（另外，Integer.MIN_VALUE == Math.abs(Integer.MIN_VALUE)爲true，負數的絕對值還是負數，這情況是不允許的）。因此，放棄這種方法。
換種另一個角度，先得出結果正負號，讓被除數，除數都是變成負數，然後讓除數左移若干位，逼近被除數，逼近得差不多，得出半成商。接着就換成用加法逼近，得出最佳近似商。最後，根據一開始保留結果正負號，得出最佳近似商正負號，並將近似商返回。
方法二 & 方法三：別人寫的，簡潔，優雅。
Submission
public class DivideTwoIntegers {
//方法一：我寫的
public int divide(int dividend, int divisor) {
if (divisor == 0)
throw new ArithmeticException();
if (dividend == Integer.MIN_VALUE && divisor == -1)
return Integer.MAX_VALUE;// 針對溢出情況
boolean negative = dividend > 0 ^ divisor > 0;
if (dividend > 0) dividend = -dividend;
if (divisor > 0) divisor = -divisor;
int quotient = 0, product = 0;
// 粗略得出商
for (int i = 0; i < 32; i++) {
int temp = divisor << i;// 相當於divisor * 2的i次方
// temp >> i != divisor, 溢出情況
if (temp < dividend || temp >> i != divisor) {
break;
} else if (temp > dividend) {
product = temp;
quotient = 1 << i;
} else {
return (negative ? -1 : 1) << i;
}
}
// 較精確逼近商
while (true) {
int temp = product + divisor;
// temp > 0, 溢出情況
if (temp < dividend || temp > 0)
break;
product = temp;
quotient++;
}
return negative ? -quotient : quotient;
}
//方法二：
public int divide2(int A, int B) {
if (A == Integer.MIN_VALUE && B == -1)
return Integer.MAX_VALUE;
int a = Math.abs(A), b = Math.abs(B), res = 0, x = 0;
while (a - b >= 0) {
for (x = 0; a - (b << x << 1) >= 0; x++)
;
res += (1 << x);
a -= (b << x);
}
return (A > 0) == (B > 0) ? res : -res;
}
//方法三：
public int divide3(int A, int B) {
if (A == Integer.MIN_VALUE && B == -1)
return Integer.MAX_VALUE;
int a = Math.abs(A), b = Math.abs(B), res = 0;
for (int x = 31; x >= 0; x--)
if ((a >>> x) - b >= 0) {
res += 1 << x;
a -= b << x;
}
return (A > 0) == (B > 0) ? res : -res;
}
}
Test
import static org.junit.Assert.*;
import org.junit.Test;
public class DivideTwoIntegersTest {
@Test
public void test() {
DivideTwoIntegers obj = new DivideTwoIntegers();
assertEquals(3, obj.divide(10, 3));
assertEquals(-2, obj.divide(7, -3));
assertEquals(0, obj.divide(0, 1));
assertEquals(1, obj.divide(1, 1));
assertEquals(Integer.MAX_VALUE, obj.divide(Integer.MAX_VALUE, 1));
assertEquals(Integer.MIN_VALUE, obj.divide(Integer.MIN_VALUE, 1));
assertEquals(0, obj.divide(1, Integer.MIN_VALUE));
assertEquals(-1, obj.divide(-1, 1));
assertEquals(1, obj.divide(Integer.MAX_VALUE, Integer.MAX_VALUE));
assertEquals(Integer.MAX_VALUE, obj.divide(Integer.MAX_VALUE, 1));
assertEquals(-Integer.MAX_VALUE, obj.divide(Integer.MAX_VALUE, -1));
assertEquals(-Integer.MAX_VALUE, obj.divide(-Integer.MAX_VALUE, 1));
assertEquals(Integer.MAX_VALUE, obj.divide(-Integer.MAX_VALUE, -1));
assertEquals(0, obj.divide(1, Integer.MAX_VALUE));
assertEquals(0, obj.divide(1, -Integer.MAX_VALUE));
for (int i = -100; i <= 100; i++) {
for (int j = 1; j < 100; j++) {
assertEquals(i / j, obj.divide(i, j));
}
}
}
@Test
public void test2() {
DivideTwoIntegers obj = new DivideTwoIntegers();
assertEquals(3, obj.divide2(10, 3));
assertEquals(-2, obj.divide2(7, -3));
assertEquals(0, obj.divide2(0, 1));
assertEquals(1, obj.divide2(1, 1));
assertEquals(Integer.MAX_VALUE, obj.divide2(Integer.MAX_VALUE, 1));
assertEquals(Integer.MIN_VALUE, obj.divide2(Integer.MIN_VALUE, 1));
assertEquals(0, obj.divide2(1, Integer.MIN_VALUE));
assertEquals(-1, obj.divide2(-1, 1));
assertEquals(1, obj.divide2(Integer.MAX_VALUE, Integer.MAX_VALUE));
assertEquals(Integer.MAX_VALUE, obj.divide2(Integer.MAX_VALUE, 1));
assertEquals(-Integer.MAX_VALUE, obj.divide2(Integer.MAX_VALUE, -1));
assertEquals(-Integer.MAX_VALUE, obj.divide2(-Integer.MAX_VALUE, 1));
assertEquals(Integer.MAX_VALUE, obj.divide2(-Integer.MAX_VALUE, -1));
assertEquals(0, obj.divide2(1, Integer.MAX_VALUE));
assertEquals(0, obj.divide2(1, -Integer.MAX_VALUE));
for (int i = -100; i <= 100; i++) {
for (int j = 1; j < 100; j++) {
assertEquals(i / j, obj.divide2(i, j));
}
}
}
@Test
public void test3() {
DivideTwoIntegers obj = new DivideTwoIntegers();
assertEquals(3, obj.divide3(10, 3));
assertEquals(-2, obj.divide3(7, -3));
assertEquals(0, obj.divide3(0, 1));
assertEquals(1, obj.divide3(1, 1));
assertEquals(Integer.MAX_VALUE, obj.divide3(Integer.MAX_VALUE, 1));
assertEquals(Integer.MIN_VALUE, obj.divide3(Integer.MIN_VALUE, 1));
assertEquals(0, obj.divide3(1, Integer.MIN_VALUE));
assertEquals(-1, obj.divide3(-1, 1));
assertEquals(1, obj.divide3(Integer.MAX_VALUE, Integer.MAX_VALUE));
assertEquals(Integer.MAX_VALUE, obj.divide3(Integer.MAX_VALUE, 1));
assertEquals(-Integer.MAX_VALUE, obj.divide3(Integer.MAX_VALUE, -1));
assertEquals(-Integer.MAX_VALUE, obj.divide3(-Integer.MAX_VALUE, 1));
assertEquals(Integer.MAX_VALUE, obj.divide3(-Integer.MAX_VALUE, -1));
assertEquals(0, obj.divide3(1, Integer.MAX_VALUE));
assertEquals(0, obj.divide3(1, -Integer.MAX_VALUE));
for (int i = -100; i <= 100; i++) {
for (int j = 1; j < 100; j++) {
assertEquals(i / j, obj.divide3(i, j));
}
}
}
}