leetcode 29. 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.

轉載自：https://blog.csdn.net/lixq05/article/details/81157304

這個思路比較清晰

class Solution {
public:
long long ABS(long long a)
{
    return a > 0 ? a : -a;
}
int divide(int dividend, int divisor)
{
    if (divisor == 0 || (dividend == INT_MIN && divisor == -1))
        return INT_MAX;
    long long a = ABS((long long)dividend);
    long long b = ABS((long long)divisor);
    long long sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
    long long res=0;
    long long tmp[33], times[33];
    //第一組數據填充
    tmp[0] = b;
    times[0] = 1;
    int index = 0;
    //一直填充到臨界大於a的位置
    while (a >= tmp[index] && index < 33)
    {
        index++;
        tmp[index] = tmp[index - 1] + tmp[index - 1];
        times[index] = times[index - 1] + times[index - 1];
    }
    //遍歷填充數據
    for (int j = index - 1; j >= 0; j--)
    {
        while (a >= tmp[j])
        {
            res += times[j];
            a -= tmp[j];
        }       
    }
    res = (sign == 1) ? res : -res;
    return (int)res;
}
};