[LeetCode] Divide Two Integers

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

 1 class Solution {
 2 private:
 3     long long f[100];
 4 public:
 5     int bsearch(long long a[], int left, int right, long long key)
 6     {
 7         if (left > right)
 8             return -1;
 9             
10         int mid = left + (right - left) / 2;
11         if (a[mid] == key)
12             return mid;
13         else if (a[mid] < key)
14         {
15             int pos = bsearch(a, mid + 1, right, key);
16             return pos == -1 ? mid : pos;
17         }
18         else
19         {
20             return bsearch(a, left, mid - 1, key);
21         }
22     }
23     
24     int divide(int dividend, int divisor) {
25         // Start typing your C/C++ solution below
26         // DO NOT write int main() function
27         int sign = dividend < 0 ? -1 : 1;
28         if (divisor < 0)
29             sign *= -1;
30         
31         long long div = dividend;
32         div = abs(div);
33         long long divisorL = divisor;
34         divisorL = abs(divisorL);
35         f[0] = divisorL;
36         int size = 1;
37         while(true)
38         {
39             if (f[size-1] >= div)
40                 break;
41             f[size] = f[size-1] + f[size-1];
42             size++;
43         }
44         
45         int num = 0;
46         long long sum = 0;
47         while(div > 0)
48         {
49             int pos = bsearch(f, 0, size - 1, div);
50             if (pos == -1)
51                 break;
52             div -= f[pos];
53             num += (1 << pos);
54         }
55                 
56         return num * sign;
57     }
58 };

 

原文地址:https://www.cnblogs.com/chkkch/p/2789800.html