[leetcode 29] Divide Two Integers

最新推荐文章于 2020-01-24 14:23:23 发布

转载最新推荐文章于 2020-01-24 14:23:23 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/ym65536/p/4268386.html

本文介绍了一种避免使用乘除和取模操作进行整数相除的方法，并详细解释了如何处理溢出情况，确保结果准确性。

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

If it is overflow, return MAX_INT.

[Solution]

使用long long避免溢出，注意dividend=0x8000000时容易产生溢出.

 1 int divide(int dividend, int divisor) 
 2     {
 3         int signf = dividend > 0 ? 1 : -1;
 4         int signs = divisor > 0 ? 1 : -1;
 5         long long ldividend = dividend > 0 ? dividend : -(long long)dividend;
 6         long long ldivisor = divisor > 0 ? divisor : -(long long)divisor;
 7         long long quotient = 0;
 8         
 9         while (ldividend >= ldivisor)
10         {
11             long long tdivisor = ldivisor;
12             long long tquotient = 1;
13             while ((tdivisor << 1) < ldividend)
14             {
15                 tdivisor <<= 1;
16                 tquotient <<= 1;
17             }
18             quotient += tquotient;
19             ldividend -= tdivisor;
20         }
21         if (signf == signs)
22             return quotient > INT_MAX ? INT_MAX : quotient;
23         else
24             return -quotient;
25     }