这道题目一直不会做，因为要考虑的corner case 太多。
1. divisor equals 0.
2. dividend equals 0.
3. Is the result negative?
4. when dividend equals Integer.MIN_VALUE and divisor equals -1, the result will overflow. convert result to long and then to integer.
5. have to use divided by divisor * 2 ^ n to avoid exceeding time limit
6. have to convert divisor and dividend to long to avoid overflow in shift.
7. constant 1 in java is compiled as integer by default.
public class Solution { public int divide(int dividend, int divisor) { if (divisor == 0) { return dividend >= 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE; } if (dividend == 0) { return 0; } boolean isNegative = (dividend > 0 && divisor < 0) || (dividend < 0 && divisor > 0); long x = Math.abs((long)dividend); long y = Math.abs((long)divisor); long result = 0; while (x >= y) { int a = 0; while (x >= (y << a)) { a++; } a--; result += (long)1 << a; x -= y << a; } if (result == (long) Integer.MAX_VALUE + 1) { return isNegative ? Integer.MIN_VALUE : Integer.MAX_VALUE; } return isNegative ? -(int)result : (int)result; } }
[leetcode] 29. divide two integers
原文：http://www.cnblogs.com/Gryffin/p/6225420.html