LeetCode | Divide Two Integers
2014-04-06 20:00
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题目
Divide two integers without using multiplication, division and mod operator.
分析
逐个减太慢,肯定超时。需要指数级增加后再减,思想类似于Linked List Cycle
II中的Brent方法。
此外,还要注意零、负数的情况。
代码
public class DivideTwoIntegers {
public int divide(int dividend, int divisor) {
if (dividend == 0 || divisor == 0) {
return 0;
}
boolean isNeg = (dividend > 0 && divisor < 0)
|| (dividend < 0 && divisor > 0);
long a = Math.abs((long) dividend);
long b = Math.abs((long) divisor);
if (b > a) {
return 0;
}
long sum = 0;
long pow = 0;
int result = 0;
while (a >= b) {
pow = 1;
sum = b;
while (sum + sum <= a) {
sum += sum;
pow += pow;
}
a -= sum;
result += pow;
}
return isNeg ? -result : result;
}
}
Divide two integers without using multiplication, division and mod operator.
分析
逐个减太慢,肯定超时。需要指数级增加后再减,思想类似于Linked List Cycle
II中的Brent方法。
此外,还要注意零、负数的情况。
代码
public class DivideTwoIntegers {
public int divide(int dividend, int divisor) {
if (dividend == 0 || divisor == 0) {
return 0;
}
boolean isNeg = (dividend > 0 && divisor < 0)
|| (dividend < 0 && divisor > 0);
long a = Math.abs((long) dividend);
long b = Math.abs((long) divisor);
if (b > a) {
return 0;
}
long sum = 0;
long pow = 0;
int result = 0;
while (a >= b) {
pow = 1;
sum = b;
while (sum + sum <= a) {
sum += sum;
pow += pow;
}
a -= sum;
result += pow;
}
return isNeg ? -result : result;
}
}
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