LeetCode 29. Divide Two Integers 二进制分解

题目

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.

题意

计算a/b的结果（计算机整数除法），不许用乘法，除法，取模计算。

如果越界，返回int的最大值

题解

初次做LeetCode，看见要算a/b,只许用加减，那应该是拿a不停减b了。范围在int的边界，如果朴素的做，来个1<<31和1瞬间就会爆。所以应该要优化一下。我们可以用二进制分解的想法来做。

a -= b

b += b

这样b的增长就会是指数级的，复杂度就会变成O(log(n))。然后用另一个变量c = 1，c+=c 来记录增长了多少次。然后减到a < b时再递归处理就可以了。

有些细节也要注意，数字有正负，同时有边界，如果不拿long long转就会wa 。

Ｃ＋＋

class Solution {
public:
int divide(int dividend, int divisor) {
int ok = 0;
long long a = dividend;
long long b = divisor;
long long div = divisor;
if (a<0 && b<0) {
a = 0LL - a;
b = 0LL - b;
div = 0 - div;
}
else if (a<0 && b >= 0) {
ok = 1;
a = 0LL - a;
}
else if (a >= 0 && b<0) {
ok = 1;
b = 0LL - b;
div = 0 - div;
}

if (a<b) return 0;

long long c = 1;
while (a >= b) {
a -= b;
b += b;
c += c;
}
//  printf("dividend = %lld divisor = %lld c = %lld\n", a, b ,c);
if (ok == 0) {
long long ans = c - 1 + divide(a, div);
if (ans == 2147483648LL)
{
return ans - 1;
}
return ans;
}
else {
long long ans = 0 - ((c - 1) + divide(a, div));
return ans;
}
}
};

看到Ｃ＋＋转来转去的太烦了。　。随手写了一个ｐｙｔｈｏｎ版的。

Python 2.7

class Solution(object):
def divide(self, dividend, divisor):
"""
:type dividend: int
:type divisor: int
:rtype: int
"""
ok = 0
if dividend < 0 and divisor < 0:
dividend = 0 - dividend
divisor = 0 - divisor
elif dividend < 0 and divisor >= 0:
dividend = 0 - dividend
ok = 1
elif dividend >= 0 and divisor < 0:
divisor = 0 - divisor
ok = 1
if dividend < divisor:
return 0
b = divisor
c = 1
while dividend >= b:
dividend = dividend - b
b = b + b
c = c + c

if ok == 0:
ans = c - 1 + Solution.divide(self, dividend, divisor)
if ans == 2147483648:
return ans - 1
else:
return ans
else:
return 0 - (c - 1 + Solution.divide(self, dividend, divisor))