[LintCode] Divide Two Integers 两数相除

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

If it is overflow, return

2147483647

Have you met this question in a real interview?

Example

Given dividend =

100

and divisor =

9

, return

11

.

LeetCode上的原题，请参见我之前的博客Divide Two Integers。

解法一：

class Solution {
public:
/**
* @param dividend the dividend
* @param divisor the divisor
* @return the result
*/
int divide(int dividend, int divisor) {
if (divisor == 0 || (dividend == INT_MIN && divisor == -1)) return INT_MAX;
long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
if (n == 1) return sign == 1 ? m : -m;
while (m >= n) {
long long t = n, p = 1;
while (m >= (t << 1)) {
t <<= 1;
p <<= 1;
}
res += p;
m -= t;
}
return sign == 1 ? res : -res;
}
};

解法二：

class Solution {
public:
/**
* @param dividend the dividend
* @param divisor the divisor
* @return the result
*/
int divide(int dividend, int divisor) {
long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
if (m < n) return 0;
while (m >= n) {
long long t = n, p = 1;
while (m > (t << 1)) {
t <<= 1;
p <<= 1;
}
res += p;
m -= t;
}
if ((dividend < 0) ^ (divisor < 0)) res = -res;
return res > INT_MAX ? INT_MAX : res;
}
};

解法三：

class Solution {
public:
/**
* @param dividend the dividend
* @param divisor the divisor
* @return the result
*/
int divide(int dividend, int divisor) {
long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
if (m < n) return 0;
long long t = n, p = 1;
while (m > (t << 1)) {
t <<= 1;
p <<= 1;
}
res += p + divide(m - t, n);
if ((dividend < 0) ^ (divisor < 0)) res = -res;
return res > INT_MAX ? INT_MAX : res;
}
};