29. Divide Two Integers
一開始選擇累加(偷懶)結果是超時,然後只好用Bit Operation,參考了一篇博文
class Solution {
public:
int divide(int dividend, int divisor) {
if (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;//同0異1
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;
}
};
相關推薦
leetcode 29. Divide Two Integers
log 線性 復雜度 viso -- iso 左移 最大 組合 數值處理的題目,有兩點要考慮的地方: 正負號的問題,對於正數除正數, 負數除正數。。。。 處理越界的問題。 解決方案:加減法,最簡單的方法是用被除數一直減去除數,直到為0, 但是復雜度高。 優化
LeetCode: 29. Divide Two Integers (Medium)
edi tro 大於 sys 復雜 ref 思路 system problems 1. 原題鏈接 https://leetcode.com/problems/divide-two-integers/description/ 2. 題目要求 給出被除數dividend和除數d
#Leetcode# 29. Divide Two Integers
https://leetcode.com/problems/divide-two-integers/ Given two integers dividend and divisor, divide two integers without using multi
[leetcode]29. Divide Two Integers
Solution 1: 轉換成long 這題太變態了 第一次碰到位運算的題。 首先判斷符號,>>>是右移運算子。負數的用補碼錶示,最高位是1,正數最高位是0。 boolean isNeg = (dividend^divisor)>>&g
【LeetCode】29. Divide Two Integers(C++)
地址:https://leetcode.com/problems/divide-two-integers/ 題目: Given two integers dividend and divisor, divide two integers without using multiplic
python leetcode 29. Divide Two Integers
注意while迴圈中的加速操作 class Solution(object): def divide(self, dividend, divisor): """ :type dividend: int :type divisor:
【LeetCode】29. Divide Two Integers - Java實現
文章目錄 1. 題目描述: 2. 思路分析: 3. Java程式碼: 1. 題目描述: Given two integers dividend and divisor, divide two integers without usi
Leetcode 29 Divide Two Integers
Given two integers dividend and divisor, divide two integers without using multiplication, division a
LeetCode 29 — Divide Two Integers(兩數相除)
Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the qu
演算法練習--LeetCode--29. Divide Two Integers
Divide Two Integers Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.Re
29. Divide Two Integers
一開始選擇累加(偷懶)結果是超時,然後只好用Bit Operation,參考了一篇博文 class Solution { public: int divide(int dividend, int divisor) { if (dividend == INT
Leetcode 29. Divide Two Integers--兩個32位整數相除,小數位截斷,不能使用乘法、除法、取模運算
Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the qu
[LeetCode]problem 29. Divide Two Integers
TAG 位操作; 二分查詢?; 除法就是被除數包含多少個除數 題目連結 方法 沒有思路,直接看得DISCUSS,寫得太好了。 首先,dividend是被除數,divisor是除數; 接著,除法表示被除數中有多少個除數,所以在不讓使用乘法、模運算的
LeetCode第29題 Divide Two Integers(兩數相除)
class Solution { public: int divide(int dividend, int divisor) { if(!divisor || (dividend == INT_MIN && divisor == -1)
LeetCode 29. 兩數相除 Divide Two Integers(C語言)
題目描述: 給定兩個整數,被除數 dividend 和除數 divisor。將兩數相除,要求不使用乘法、除法和 mod 運算子。 返回被除數 dividend 除以除數 divisor 得到的商。 示例 1: 輸入: dividend = 10, divisor =
leetcode 29. 兩數相除(Divide Two Integers)
code cpp lis 示例 vid int else divide tle 目錄 題目描述: 示例 1: 示例 2: 解法: 題目描述:
LeetCode 28 Divide Two Integers
範圍 long mod max article edi 優化 指數 故障 Divide two integers without using multiplication, division and mod operator. 思路:1.先將被除數和除數轉化為long的
Divide Two Integers
btn button value param pan logs question math multipl Divide two integers without using multiplication, division and mod operator. If i
[Swift]LeetCode29. 兩數相除 | Divide Two Integers
Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient a
LeetCode Day25 Divide Two Integers
10/3 ==> 最接近10的數:3* 21+3* 20(divisor* 21+divisor* 20)所以商為21+20=3 class Solution { public: int divide(int dividend, int divisor) {