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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.
考察:int类型两数整除结果,转化为高阶long long类型来运算,如果溢出则返回INT_MAX, 尽可能去使用位移动,以及异或运算求符号;
class Solution {
public:
int divide(int dividend, int divisor) {
long long res = 0;
long long m = abs((long long)dividend), n = abs((long long)divisor);
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)) //符号取异或
res = -res;
return res > INT_MAX ? INT_MAX : res; // 判断是否溢出
}
};
完,