Given two integers dividend
and divisor
, divide two integers without using multiplication, division and mod operator.
Return the quotient after dividing dividend
by divisor
.
The integer division should truncate toward zero.
Example 1:
Input: dividend = 10, divisor = 3
Output: 3
Example 2:
Input: dividend = 7, divisor = -3
Output: -2
Note:
- Both dividend and divisor will be 32-bit signed integers.
- The divisor will never be 0.
- Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 231 − 1 when the division result overflows.
这道题让我们求两数相除,而且规定不能用乘法,除法和取余操作,那么这里可以用另一神器位操作 Bit Manipulation,思路是,如果被除数大于或等于除数,则进行如下循环,定义变量t等于除数,定义计数p,当t的两倍小于等于被除数时,进行如下循环,t扩大一倍,p扩大一倍,然后更新 res 和m。这道题的 OJ 给的一些 test case 非常的讨厌,因为输入的都是 int 型,比如被除数是 -2147483648,在 int 范围内,当除数是 -1 时,结果就超出了 int 范围,需要返回 INT_MAX,所以对于这种情况就在开始用 if 判定,将其和除数为0的情况放一起判定,返回 INT_MAX。然后还要根据被除数和除数的正负来确定返回值的正负,这里采用长整型 long 来完成所有的计算,最后返回值乘以符号即可,代码如下:
解法一:
class Solution {
public:
int divide(int dividend, int divisor) {
if (dividend == INT_MIN && divisor == -) return INT_MAX;
long m = labs(dividend), n = labs(divisor), res = ;
int sign = ((dividend < ) ^ (divisor < )) ? - : ;
if (n == ) return sign == ? m : -m;
while (m >= n) {
long t = n, p = ;
while (m >= (t << )) {
t <<= ;
p <<= ;
}
res += p;
m -= t;
}
return sign == ? res : -res;
}
};
我们可以通过递归的方法来解使上面的解法变得更加简洁:
解法二:
class Solution {
public:
int divide(int dividend, int divisor) {
long m = labs(dividend), n = labs(divisor), res = ;
if (m < n) return ;
long t = n, p = ;
while (m > (t << )) {
t <<= ;
p <<= ;
}
res += p + divide(m - t, n);
if ((dividend < ) ^ (divisor < )) res = -res;
return res > INT_MAX ? INT_MAX : res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/29
参考资料:
https://leetcode.com/problems/divide-two-integers/
https://leetcode.com/problems/divide-two-integers/discuss/13524/summary-of-3-c-solutions
https://leetcode.com/problems/divide-two-integers/discuss/13407/C%2B%2B-bit-manipulations