LeetCode----29. Divide Two Integers (两数相除)

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.

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//不能使用乘法，或者除法
//考虑除数和被除数为零的情况

Input:-2147483648-1

Output:0

Expected:2147483647
Math.abs(int)-----超出int范围，还是负数

Input:-21474836482

Output:0

Expected:-1073741824

Submission Result: Time Limit Exceeded  More
Details 

Last executed input:-21474836482

public class Solution {
public int divide(int dividend, int divisor) {
int sum=0;

if(divisor==0){return dividend>=0?Integer.MAX_VALUE:Integer.MIN_VALUE;}
if ((dividend == Integer.MIN_VALUE && divisor == -1)||(dividend == Integer.MAX_VALUE && divisor == 1)) {
return Integer.MAX_VALUE;
}
if ((dividend == Integer.MAX_VALUE && divisor == -1)||(dividend == Integer.MIN_VALUE && divisor == 1)) {
return Integer.MIN_VALUE;
}
if((dividend>0&&divisor<0)||(dividend<0&&divisor>0)){
long dd=Math.abs((long)dividend);long ds=Math.abs((long)divisor);
while(dd>=ds){dd=dd-ds;sum++;}
return 0-sum;}
else {long dd=Math.abs((long)dividend);long ds=Math.abs((long)divisor);
while(dd>=ds){dd=dd-ds;sum++;}
return sum;}
}

}

、、\\泰勒公式

public class Solution {
/**
* @param dividend the dividend
* @param divisor the divisor
* @return the result
*/
public int divide(int dividend, int divisor) {
if (divisor == 0) {
return dividend >= 0? Integer.MAX_VALUE : Integer.MIN_VALUE;
}

if (dividend == 0) {
return 0;
}

if (dividend == Integer.MIN_VALUE && divisor == -1) {
return Integer.MAX_VALUE;
}

boolean isNegative = (dividend < 0 && divisor > 0) ||
(dividend > 0 && divisor < 0);

long a = Math.abs((long)dividend);
long b = Math.abs((long)divisor);
int result = 0;
while(a >= b){
int shift = 0;
while(a >= (b << shift)){
shift++;
}
a -= b << (shift - 1);
result += 1 << (shift - 1);
}
return isNegative? -result: result;
}
}