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Quesiton from: https://leetcode.com/problems/divide-two-integers/
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
面试里,尽量不要使用 long 来处理 Integer Overflow。因为这道题目本身考查的就是各种 Corner Cases 的处理。
1 public int divide(int dividend, int divisor) { 2 if(divisor == 0) { 3 return Integer.MAX_VALUE; 4 } 5 6 if(dividend == 0) { 7 return 0; 8 } 9 10 if(divisor == Integer.MIN_VALUE) { 11 return (dividend == divisor)?1:0; 12 } 13 14 int sign = (((dividend^divisor) >> 31) == 0)? 1:-1; 15 int res = 0; 16 int num1 = 0; 17 int num2 = Math.abs(divisor); 18 19 20 if(dividend == Integer.MIN_VALUE) { 21 if(divisor == -1) { 22 return Integer.MAX_VALUE; 23 } else { 24 res = 1; 25 num1 = Math.abs(dividend + num2); 26 } 27 } else { 28 num1 = Math.abs(dividend); 29 } 30 31 res = helper(num1, num2, res); 32 33 return (sign>0)?res:-res; 34 } 35 36 private int helper(int num1, int num2, int res) { 37 int times = 1; 38 39 while(num2 <= (num1>>1)) { 40 num2 = num2<<1; 41 times = times<<1; 42 } 43 44 while(times > 0) { 45 if(num1 >= num2) { 46 num1 -= num2; 47 res += times; 48 } 49 times >>= 1; 50 num2 >>= 1; 51 } 52 53 return res; 54 }
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原文地址:http://www.cnblogs.com/Phoenix-Fearless/p/5097693.html