**[Lintcode]Divide Two Integers
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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
.
不能使用乘除和取模,这时可以考虑使用位运算。每次将divisor左移一位,直到超过dividend。为了便于计算,统一将两个数字转换为负数。
例子中9可以左移3位,即9×2×2×2=72,再左移的话就会超过dividend。这时重新计算dividend = 100 - 72 = 28.
再次左移,9×2=18,重新计算dividend=28-18=10.
再次左移,9×1 = 9,重新计算divident = 10-9=1,小于divisor,停止。
最后结果为2×2×2+2+1=11
需要注意的情况:2147483647, 1,-2147483648, 1,-2147483648, -2147483648
divisor为1时,计算耗时,可以单独考虑。divisor<<1就越界时,可以单独考虑。重点考虑越界情况。
public class Solution { /** * @param dividend the dividend * @param divisor the divisor * @return the result */ public int divide(int dividend, int divisor) { //2147483647, 1, when divisor = 1, m can depass the limit and get neg value boolean neg = false; if((dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0)) neg = true; if(divisor == 1) return dividend; if(divisor == -1) return dividend == Integer.MIN_VALUE ? Integer.MAX_VALUE : -dividend; if(dividend > 0) dividend = -dividend; if(divisor > 0) divisor = -divisor; if(divisor << 1 >= 0) { if(dividend <= divisor) return neg ? -1 : 1; else return 0; } int res = 0, k = dividend; while(k <= divisor) { int tmp = divisor; int m = 1; while(tmp >= k && tmp << 1 < 0) { tmp = tmp << 1; m = m << 1; } k = k - (tmp >> 1); res = res + (m >> 1); } return neg ? -res : res; }}
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