标签：
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return 2147483647
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
1 public class Solution { 2 /** 3 * @param dividend the dividend 4 * @param divisor the divisor 5 * @return the result 6 */ 7 public int divide(int a, int b) { 8 9 if (b == 0 || (a == -2147483648 && b == -1)) return 2147483647; 10 if (b == 1) return a; 11 if (b == -1) return 0 - a; 12 13 int count = 0; 14 for(long i = abs(a); i >= abs(b); i -= abs(b)) { 15 count++; 16 } 17 18 if (positive(a, b) == true) 19 return count; 20 else 21 return 0 - count; 22 } 23 24 public boolean positive(int a, int b) { 25 if ((a > 0 && b > 0) || (a < 0 && b < 0)) 26 return true; 27 else 28 return false; 29 } 30 31 public long abs(long value) { 32 if (value >= 0 ) 33 return value; 34 else 35 return 0 - value; 36 } 37 }
在leetcode里进行测试，发现超时。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 /** 3 * @param dividend the dividend 4 * @param divisor the divisor 5 * @return the result 6 */ 7 public int divide(int dividend, int divisor) { 8 if (divisor == 0) { 9 return dividend >= 0? Integer.MAX_VALUE : Integer.MIN_VALUE; 10 } 11 12 if (dividend == 0) { 13 return 0; 14 } 15 16 if (dividend == Integer.MIN_VALUE && divisor == -1) { 17 return Integer.MAX_VALUE; 18 } 19 20 boolean isNegative = (dividend < 0 && divisor > 0) || 21 (dividend > 0 && divisor < 0); 22 23 long a = Math.abs((long)dividend); 24 long b = Math.abs((long)divisor); 25 int result = 0; 26 while(a >= b){ 27 int shift = 0; 28 while(a >= (b << shift)){ 29 shift++; 30 } 31 a -= b << (shift - 1); //剩余部分 32 result += 1 << (shift - 1); // 1 << (shift - 1) = (b << (shift - 1)) / b 33 } 34 return isNegative? -result: result; 35 } 36 }
Reference:
http://www.jiuzhang.com/solutions/divide-two-integers/
标签：
原文地址：http://www.cnblogs.com/beiyeqingteng/p/5668465.html
