[leetcode]29. Divide Two Integers两整数相除

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3

Example 2:

Input: dividend = 7, divisor = -3
Output: -2

Note:

• Both dividend and divisor will be 32-bit signed integers.
• The divisor will never be 0.
• Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−  2³¹,  2³¹ − 1]. For the purpose of this problem, assume that your function returns 2³¹ − 1 when the division result overflows.

题意：

不准乘除，不准膜。

Solution1：用加减和位运算。

乘法可以通过不断加法来获得

故，除法可以通过不断减法来获得

碎碎念：

数学渣对这种没有算法含量，又要大量数学sense的题，好抓狂。

发挥文科生优势。背诵吧！！

code

 class Solution {
     public int divide(int dividend, int divisor) {
         // corner case
         if(dividend == 0) return 0;         // 当 dividend = INT_MIN，divisor = -1时，结果会溢出
         if (dividend == Integer.MIN_VALUE) {
             if (divisor == -1) return Integer.MAX_VALUE;
             else if (divisor < 0)
                 return 1 + divide( dividend - divisor, divisor);
             else
                 return - 1 + divide( dividend + divisor, divisor);
         }         if(divisor == Integer.MIN_VALUE){
             return dividend == divisor ? 1 : 0;
         }         int a = dividend > 0 ? dividend : -dividend;
         int b = divisor > 0 ? divisor : -divisor;         int result = 0;
         while (a >= b) {
             int c = b;
             for (int i = 0; a >= c;) {
                 a -= c;
                 result += 1 << i;
                 if (c < Integer.MAX_VALUE / 2) { // prevent overflow
                     ++i;
                     c <<= 1;
                 }
             }
         }         return ((dividend^divisor) >> 31) != 0 ? (-result) : (result);
     }
 }