题意：

image.png

解题思路：

  1. 思路：
  2.     1. 模拟 + 倍增
  3.     2. 标记正负号，然后转成绝对值进行操作
  4.     3. dividend=11，divisor=3
  5.     4. (3 + 3 = 6 < 11) => 得出最少有2个3，所以倍数为2
  6.     5. 6 + 6 > 11,但是11 - 6 = 5，5大于3，得出最少可得1个3

PHP代码实现：

  1. class Solution {
  3.     function divide($dividend, $divisor) {
  4.         $sign = 1;
  5.         if (($dividend > 0 && $divisor < 0) 
  6.             || ($dividend < 0 && $divisor > 0)) $sign = -1;
  8.         $ldivedend = abs($dividend);
  9.         $ldivisor = abs($divisor);
  11.         if ($ldivedend < $ldivisor || $ldivedend == 0) return 0;
  13.         $lres = $this->div($ldivedend, $ldivisor);
  14.         $num = $sign < 1 ? 0 - $lres : $lres;
  15.         if ($num > pow(2, 31) - 1) return pow(2, 31) - 1;
  16.         if ($num < pow(-2, 31)) return pow(-2, 31);
  18.         return $num;
  19.     }
  21.     function div($ldivedend, $ldivisor) {
  22.         if ($ldivedend < $ldivisor) return 0;
  24.         $sum = $ldivisor;
  25.         $mult = 1;
  27.         while (($sum + $sum) < $ldivedend) {
  28.             $sum += $sum;
  29.             $mult += $mult;
  30.         }
  31.         return $mult + $this->div($ldivedend - $sum, $ldivisor);
  32.     }
  33. }

GO代码实现：

  1. func divide(dividend int, divisor int) int {
  3.     if dividend == math.MinInt32 && divisor == -1 {
  4.         return math.MaxInt32
  5.     }
  7.     sign := 1
  8.     if (divisor > 0 && dividend < 0) || (divisor < 0 && dividend > 0) {
  9.         sign = -1
  10.     }
  11.     if divisor < 0 { divisor *= -1 }
  12.     if dividend < 0 { dividend *= -1 }
  13.     res := div(int64(dividend), int64(divisor))
  15.     if sign == -1 {
  16.         return -res
  17.     }
  18.     return res
  19. }
  21. func div(a int64,b int64) int {
  22.     if a < b {
  23.         return 0
  24.     }
  25.     sum := b
  26.     mul := 1
  28.     for (sum + sum) < a {
  29.         sum += sum
  30.         mul += mul
  31.     }
  32.     return mul + div(a - sum, b)
  33. }