Divide two integers without using multiplication, division and mod operator.
1 class Solution { 2 public: 3 int divide(int dividend, int divisor) { 4 long long a = dividend < 0 ? -(long long)dividend : dividend; 5 long long b = divisor < 0 ? -(long long)divisor : divisor; 6 bool sign = (dividend >= 0 && divisor >= 0) || 7 (dividend <= 0 && divisor <= 0); 8 long long ans = 0; 9 while (a >= b) { 10 long long c = b; 11 for (int i = 0; c <= a; c = c << 1, ++i) { 12 a -= c; 13 ans += (1 << i); 14 } 15 } 16 return sign ? ans : -ans; 17 } 18 };
不能用乘、除和取模,那么还可以用加、减和移位。
符号单独考虑。主要从减法出发,可以通过移位加倍被除数从而实现加速。有一点要特别注意的是可能发生溢出,需要用long long类型来进行中间的运算。
【Leetcode】Divide Two Integers,布布扣,bubuko.com
原文地址:http://www.cnblogs.com/dengeven/p/3766368.html