Leetcode 29 :Divide two integers Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT...LeetCode 29 Divide Two Integers Created with Raphaël 2.1.0开始遍历[0,31],查看b是否为2的次幂确认?返回ans=a<<log b结束二分查找区间[a>>(x...
int divide(int dividend, intdivisor) { if(divisor == 0)return MAX_INT;//除数为零 if(divisor == -1 && dividend == -2147483648)return MAX_INT;//被除数为int的最小数(-2147483648),除数为-1,此时由于int整数最大只能为2147483647,会溢出 int flag = 1; if(dividend > 0){//计算符号 dividend...
LeetCode 29. Divide Two Integers Description Given two integers dividend and divisor, divide two integers without using multiplication, division, and mod operator. The integer division should truncate toward zero, which means losing its fractional part. For example, 8.345 would be truncated to 8, a...
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...
Leetcode: Divide Two Integers Divide two integers without using multiplication, division and mod operator. Best method(跟discuss vote最高相似):只有一种情况整数除以整数会overflow,那就是Integer.MIN_VALUE除以-1,这种情况特殊分析。 之所以要用long a, b代替dividend和divisor是因为:比如Integer.MAX_VALUE除以...
LeetCode-Divide Two Integers Description: 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....
2. 还要考虑一种特殊的corner case: dividend =Integer.MIN_VALUE, divisor = -1, 此时res =Integer.MAX_VALUE +1,所以cast时会出错,要特殊处理. 3. 当以2的幂次方增长时,就用如下方式来写: 1while(){2sum+=sum;3pow+=pow;4} 1. 2.
Divide Two Integers -- LeetCode 原题链接:http://oj.leetcode.com/problems/divide-two-integers/ 这道题属于数值处理的题目,对于整数处理的问题,在Reverse Integer中我有提到过,比较重要的注意点在于符号和处理越界的问题。对于这道题目,因为不能用乘除法和取余运算,我们只能使用位运算和加减法。比较直接的方法...
[LeetCode]--29. Divide Two Integers 简介:Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT.以前我记得做过乘法变加法吧,这个有点像除法变减法,用位运算,二进制嘛,左移一位相当于乘以二。一个有趣的是 Math.abs(-2147483648...
if (res >= Integer.MAX_VALUE) res = neg ? Integer.MIN_VALUE : Integer.MAX_VALUE; else if (neg) res = -res; return (int)res; } } 看得懂的版本: public class Solution { public int divide(int dividend, int divisor) { boolean isNeg = (dividend < 0 && divisor > 0) || (divide...