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.

這個題的意思是在不使用乘除和取模運算的情況下進行除運算。

1）

class Solution {
    public int divide(int dividend, int divisor) {
        if(divisor == 0 || (dividend == Integer.MIN_VALUE && divisor == -1)){
            return Integer.MAX_VALUE;
        }
        int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;//判斷兩個數中是否存在負數，存
        在的話結果為false，不存在的話為true
        long m = Math.abs((long)dividend);//直接對正整數進行運算
        long n = Math.abs((long)divisor);
        long res = 0;
        if(n == 1){
            return sign == -1 ? (int)m * -1 : (int)m;//標記正負號
        }
        while(m >= n){
            long t = n,sum = 1;
            while(m >= (t << 1)){//進行簡便計算
                t <<= 1;
                sum <<= 1;
            }
            res += sum;
            m -= t;
        }
        return sign == -1 ? (int)res * -1 : (int)res;//最後的時候將符號加到原來的值上面。
    }
}

2）

class Solution {
    public int divide(int dividend, int divisor) {
       long a = Math.abs((long)dividend);
         long b = Math.abs((long)divisor);
         
         long ret = 0;
         
         while (a >= b) {
             for (long tmp = b, cnt = 1; a >= tmp; tmp <<= 1, cnt <<= 1) {
                 ret += cnt;
                 a -= tmp;
             }
         }
         
         ret = (((dividend ^ divisor) >> 31) & 1) == 1 ? -ret: ret;
         if (ret > Integer.MAX_VALUE || ret < Integer.MIN_VALUE) {
             return Integer.MAX_VALUE;
         }         
       return (int)ret;
    }
}

使用了位的表示法（32位）。