[LeetCode] Divide Two Integers 两数相除 Divide two integers without using multiplication, division and mod operator. If it is overflow, retu ... [LeetCode] 29. Divide Two Integers 两数相除 Given two integers dividen
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 231−1when the division result overflows. 题意:实现两个...
Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT链接https://leetcode.com/problems/divide-two-integers/答案1、int的最大值MAX_INT为power(2,31)-1 = 21474836472、int的最小值MIN_INT为-power(2,31) = -21474836483、当MIN_INT除以-1的...
代码如下: 1publicclassSolution {2publicintdivide(intdividend,intdivisor) {3//被除数为0或者Integer.MIN_VALUE / -1的情况。4if(divisor == 0 || (dividend ==Integer.MIN_VALUE5&& divisor == -1))returnInteger.MAX_VALUE;6intres = 0;7//符号8intsign = (dividend > 0 && divisor < 0) || ...
Leetcode每日一题:29.divide-two-integers(两数相除),参照评论大佬思路学到了:/***解题思路:这题是除法,所以先普及下除法术语*商,公式是:(被除数-余数)÷除数=商,记作:被除数÷除数=商...余数,是一种数学术语。*在一个除法算式里,被除数、余数、除数和商的关系为
题目链接: Divide Two Integers : https://leetcode.com/problems/divide-two-integers/ 两数相除: https://leetcode.cn/problems/divide-two-integers/ LeetCode 日更第136天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满
所以4 + (- 2)= 4 + 14 = 2。 我们利用同余的性质,把减法成功转换成了加法,所以我们只需要在计算机里边将 -2 存成 14 就行了。我们这里减去 2 就等价于加上 14。 再比如 13 - 7 ,也就是 13 + (-7) 13 ≡ 13 (mod 16) -7 ≡ 9(mod 16) ...
6 在最开始的时候,我们需要判断dividend和divisor 7 positive = (dividend<0) is (divisor<0) 这句code作用是:如果is两端都为True的话,positive就是True;如果is两端都是False的话,positive也是True;两个一个为正,一个为负的话,positive就是False 8 a << 1 is equal to a * 2...
with an environment that could only store integers within the 32-bit signed integer range: −2^31, 2^31 − 1. For this problem, if the quotient is strictly greater than 2^31 - 1, then return 2^31 - 1, and if the quotient is strictly less than -2^31, then return -2^31. ...
leetcode 29 Divide Two Integers 题目详情 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 题目要求我们在不借助乘法运算、除法运算和模运算的基础上,求出输入的两个整数相除的结果。如果溢出,那么返回MAX_INT。其中第一个参数是被除数,第二个...