leetcode -- Divide Two Integers
Divide two integers without using multiplication, division and mod operator.
1.naive的做法是用除数一个一个减被除数，直到被除数<=0, 此方法会超时！！！
2.将被除数向左移位(x2),然后与被除数做比较，直到发现仅次于被除数的那个值，减去该值后继续(line:24-30)。
1 public class Solution { 2 public static int divide(int dividend, int divisor) { 3 if (dividend == 0 || divisor == 1) 4 return dividend; 5 6 int sign = 1; 7 if (dividend < 0) 8 sign = sign * -1; 9 if (divisor < 0) 10 sign = sign * -1; 11 12 long divid = Math.abs((long)dividend); 13 long divis = Math.abs((long)divisor); 14 15 long[] a = new long[32]; 16 int i = 0; 17 while (divis > 0 && divis <= divid) { 18 a[i] = divis; 19 divis = divis << 1; 20 i++; 21 } 22 i--; 23 long res = 0; 24 while ( divid > 0 && i >= 0) { 25 if(divid >= a[i]){ 26 divid = divid - a[i]; 27 res = res + (1 << i); 28 } 29 i--; 30 } 31 return (int)(res * sign) ; 32 } 33 }
使用int时会出现整数溢出问题，INT_MIN(-2147483648)，比如 abs(-2147483648) = -2147483648，这里在12、13行全部使用long
另外+的优先级要比<<高
ref:http://fisherlei.blogspot.com/2012/12/leetcode-divide-two-integers.html
