29. Divide Two Integers (Medium)
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.
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: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 231 − 1 when the division result overflows.
Solutions
class Solution {
/*
* This problem is really frustrating with a mass of corn cases.
* Interviewers much take good care of overflow issue. To get rid of overflow,
* It's a good habit to convert all the variables to long type before calculation.
*/
public int divide(int dividend, int divisor) {
// Be careful of the overflow issue
if (dividend == Integer.MIN_VALUE && divisor == -1) {
return Integer.MAX_VALUE;
}
if (divisor == 1) {
return dividend;
}
if (divisor == -1) {
return -dividend;
}
boolean isNegative = ((dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0));
// do not use below statements, Math.abs can be handle int overflow issue
// long tmpDvd = Math.abs(dividend);
// long tmpDvs = Math.abs(divisor);
long tmpDvd = dividend;
long tmpDvs = divisor;
tmpDvd = Math.abs(tmpDvd);
tmpDvs = Math.abs(tmpDvs);
int ans = 0;
long remain = tmpDvd;
while (true) {
if (remain - tmpDvs < 0) {
break;
}
int count = 0;
// 1. comparision must be >= 0, not > 0
// 2. tmpDvs must be long type, otherwise overflow takes place
while (remain - (tmpDvs << count) >= 0) {
count++;
}
count--;
// to reach here, count must be >= 0
ans += (1 << count);
// notice remain - divisor << count is absolutely different from remain - (divisor << count)
remain -= tmpDvs << count;
}
return isNegative ? -ans : ans;
}
}