Leetcode 29 Divide Two Integers

最新推荐文章于 2020-01-24 14:23:23 发布

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最新推荐文章于 2020-01-24 14:23:23 发布

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本文链接：https://blog.youkuaiyun.com/Neo233/article/details/83420235

本文介绍了一种在不使用乘除和取模运算的情况下进行除法运算的方法，适用于32位整数，通过位操作实现高效计算，同时处理正负数情况。

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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.

这个题的意思是在不使用乘除和取模运算的情况下进行除运算。

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;//判断两个数中是否存在负数，存
        在的话结果为false，不存在的话为true
        long m = Math.abs((long)dividend);//直接对正整数进行运算
        long n = Math.abs((long)divisor);
        long res = 0;
        if(n == 1){
            return sign == -1 ? (int)m * -1 : (int)m;//标记正负号
        }
        while(m >= n){
            long t = n,sum = 1;
            while(m >= (t << 1)){//进行简便计算
                t <<= 1;
                sum <<= 1;
            }
            res += sum;
            m -= t;
        }
        return sign == -1 ? (int)res * -1 : (int)res;//最后的时候将符号加到原来的值上面。
    }
}

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）。