29. Divide Two Integers
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return 2147483647
Example
Given dividend = 100 and divisor = 9, return 11.
分析：
这本身并不难，难的是里面有很多想不到的test case。比如：
a = -2147483648 && b = -1， 应该返回 2147483647
a = -2147483648 && b = 2 应该返回 -1073741824
这题的终极解法应该是用bit operators. 原理是快速找出最大的以divisor为底数的值，当然这值必须要小于dividend。假设那个值是K， 那么我们可以确定。 dividend / divisor = K / divisor + (dividend - K) / divisor.
如何快速找出最大的以divisor为底数的值呢，用 << Operator.
1 public class Solution { 2 public int divide(int dividend, int divisor) { 3 if (dividend == 0) return 0; 4 if (divisor == 1) return dividend; 5 if (dividend == Integer.MIN_VALUE && divisor == -1) return Integer.MAX_VALUE; 6 7 boolean isNegative = (dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0); 8 // case: -2147483648, 2 9 long a = Math.abs((long) dividend), b = Math.abs((long) divisor); 10 11 int result = 0; 12 // if we divide 100 by 5 by using for loop, we need to iterate for 20 times (answer). Can we make it shorter? 13 // The answer is we can increase the answer by 2 times each time until divisor * answer > dividend. 14 while (a >= b) { 15 int shift = 0; 16 for (; a >= (b << shift); shift++); 17 a -= b << (shift - 1); // 剩余部分 18 result += 1 << (shift - 1); 19 } 20 return isNegative ? -result : result; 21 } 22 }
Reference:
http://www.jiuzhang.com/solutions/divide-two-integers/
