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# 29.Divide-Two-Integers

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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, which means losing its fractional part. For example, truncate(8.345) = 8 and truncate(-2.7335) = -2.
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Example 1:
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Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = truncate(3.33333..) = 3.
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Example 2:
Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = truncate(-2.33333..) = -2.
Note:
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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: [â231, 231 â 1]. For the purpose of this problem, assume that your function returns 231 â 1 when the division result overflows.

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### Approach #1 Repeated Subtraction

public int divide(int dividend, int divisor) {
//Reduce the problem to positive long integer to make it easier.
//Use long to avoid integer overflow cases.
int sign = 1;
if ((dividend > 0 && divisor < 0) || (dividend < 0 && divisor > 0))
sign = -1;
long ldividend = Math.abs((long) dividend);
long ldivisor = Math.abs((long) divisor);
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//Take care the edge cases.
if (ldivisor == 0) return Integer.MAX_VALUE;
if ((ldividend == 0) || (ldividend < ldivisor)) return 0;
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long lans = ldivide(ldividend, ldivisor);
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int ans;
if (lans > Integer.MAX_VALUE){ //Handle overflow.
ans = (sign == 1)? Integer.MAX_VALUE : Integer.MIN_VALUE;
} else {
ans = (int) (sign * lans);
}
return ans;
}
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private long ldivide(long ldividend, long ldivisor) {
// Recursion exit condition
if (ldividend < ldivisor) return 0;
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// Find the largest multiple so that (divisor * multiple <= dividend),
// whereas we are moving with stride 1, 2, 4, 8, 16...2^n for performance reason.
// Think this as a binary search.
long sum = ldivisor;
long multiple = 1;
while ((sum+sum) <= ldividend) {
sum += sum;
multiple += multiple;
}
//Look for additional value for the multiple from the reminder (dividend - sum) recursively.
return multiple + ldivide(ldividend - sum, ldivisor); // 4 + 2 + 1
}

### Approach #2

This solution has `O(logN^2)` time complexity.
public int divide(int A, int B) {
if (A == 1 << 31 && B == -1) {
return (1 << 31) - 1;
}
int a = Math.abs(A), b = Math.abs(B);
int res = 0, x = 0;
while (a - b >= 0) {
for (x = 0; a - (b << x << 1) >= 0; x++);
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res += 1 << x;
a -= b << x;
}
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return (A > 0) == (B > 0) ? res: -res;
}