LeetCode Online Judge 题目C# 练习 - Divid Two Integer

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

 1 public static int Divid(int dividend, int divisor)

 2         {

 3             bool neg = false;

 4             uint uint_dividend;

 5             uint uint_divisor;

 6 

 7             if (dividend < 0)

 8             {

 9                 uint_dividend = (uint)~dividend + 1; // 等同于Convert.ToUInt32(dividend);

10                 neg = !neg;

11             }

12             else

13                 uint_dividend = (uint)dividend;

14 

15             if (divisor < 0)

16             {

17                 uint_divisor = (uint)~divisor + 1;

18                 neg = !neg;

19             }

20             else

21                 uint_divisor = (uint)divisor;

22 

23             int quotient = 0;

24 

25             while(uint_dividend > divisor) //每次被除数-除数，商++

26             {

27                 quotient++;

28                 uint_dividend -= uint_divisor;

29             }

30 

31             return neg ? -quotient : quotient;

32         }

代码分析：

　　这种方便比较笨，就是简单的一次一次循环被除数-除数，然后计算减了多少次，INT32.MaxValue / 1 就会让你哭了，虽然可以加以特殊条件，当除数 == 1，返回被除数，但是 INT32.MaxValue / 2 也够好受的。

 1 public static int DividOpt(int dividend, int divisor)

 2         {

 3             bool neg = false;

 4             uint uint_dividend;

 5             uint uint_divisor;

 6 

 7             if (dividend < 0)

 8             {

 9                 uint_dividend = (uint)~dividend + 1; // 等同于Convert.ToUInt32(dividend);

10                 neg = !neg;

11             }

12             else

13                 uint_dividend = (uint)dividend;

14 

15             if (divisor < 0)

16             {

17                 uint_divisor = (uint)~divisor + 1;

18                 neg = !neg;

19             }

20             else

21                 uint_divisor = (uint)divisor;

22 

23             uint ret = div(uint_dividend, uint_divisor);

24 

25             if (neg)

26                 return (int)~ret + 1;

27 

28             return (int)ret;

29         }

30 

31         public static uint div(uint dividend, uint divisor)

32         {

33             if (dividend < divisor)

34                 return 0;

35             if (dividend == divisor)

36                 return 1;

37 

38             if (dividend > (divisor << 1) && (divisor << 1) > divisor) //如果divisor右移1位 < dividend 而且不溢出

39             {

40                 uint temp = divisor << 1;

41                 int bit_num = 1; //右移位数计数器

42                 while (dividend > (temp << 1))

43                 {

44                     bit_num++;

45                     if ((temp << 1) > temp) //保证临时除数temp不溢出

46                         temp <<= 1;

47                     else

48                         break;

49                 }

50 

51                 return div(dividend - temp, divisor) + (uint)(1 << bit_num); // 例如(100 - 64 ， 2) + 32， 32  = 64 / 2;             }

53             else

54                 return div(dividend - divisor, divisor) + 1;

55         }

 

代码分析：

　　这个优化版本利用了移位，让算法从O(dividen / divisor）提升到了O（lg(dividen / divisor))

　　主要算法是，循环做divisor << 1 到 > dividen之前的一个存到temp，然后递归 div( dividen - temp, divisor) + （1 << bit_num）