LeetCode(29)Divide Two Integers
来源:互联网 发布:新东方网络学校 编辑:程序博客网 时间:2024/05/18 00:11
题目
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
分析
题目要求不用 * / %三种运算符的条件下,求得两个int类型整数的商。
方法一:
很明显的,我们可以用求和累计的方法,求得商,但是该方法测试会出现TLE;参考博客提出解决办法:每次将被除数增加1倍,同时将count也增加一倍,如果超过了被除数,那么用被除数减去当前和再继续本操作,但是我测试结果依然是TLE。所以这道题的目的在于考察逻辑运算。
方法二:
该方法来源于参考博客但是该实现忽略了结果溢出的问题,需要加上结果是否溢出判断。
TLE(方法一)代码
//方法一,翻倍累和 结果是:Time Limit Exceededclass Solution {public: int divide(int dividend, int divisor) { //如果被除数或者除数有一者为0 或者绝对值除数大于被除数则返回0 if (dividend == 0 || divisor == 0 || abs(divisor) > abs(dividend)) return 0; int sign = ((dividend > 0 && divisor > 0) || (dividend < 0 && divisor < 0)) ? 1 : -1; long long Dividend = abs(dividend), Divisor = abs(divisor); long long sum = 0; int count = 0, ret = 0; while (Divisor <= Dividend) { count = 1; sum = Divisor; while ((sum + sum) < Dividend) { sum += sum; count += count; } Dividend -= sum; ret += count; } if (sign == -1) return 0 - ret; else return ret; }};
AC代码
//方法二:位运算class Solution {public: int divide(int dividend, int divisor) { //如果被除数或者除数有一者为0 或者绝对值除数大于被除数则返回0 if (dividend == 0 || divisor == 0) return 0; // without using * / mod // using add auto sign = [=](long long x) { return x < 0 ? -1 : 1; }; int d1 = sign(dividend); int d2 = sign(divisor); long long n1 = abs(static_cast<long long>(dividend)); long long n2 = abs(static_cast<long long>(divisor)); long long ans = 0; while (n1 >= n2) { long long base = n2; for (int i = 0; n1 >= base; ++i) { n1 -= base; base <<= 1; ans += 1LL << i; } } //如果转换为int类型,结果溢出,返回INT_MAX ,int类型表示范围[-2147483648 , 2147483648) if (ans > INT_MAX && d1 == d2) return INT_MAX; int res = static_cast<int>(ans); if (d1 != d2) return -res; else return res; }};
GitHub测试程序源码
0 0
- 【leetcode-29】Divide Two Integers(C)
- LeetCode(29)Divide Two Integers
- LeetCode (29)Divide Two Integers
- leetcode 29: Divide Two Integers
- LeetCode(29)Divide Two Integers
- [leetcode 29] Divide Two Integers
- [leetcode] 29 Divide Two Integers
- leetcode 29 Divide Two Integers
- Leetcode 29 Divide Two Integers
- LeetCode 29 - Divide Two Integers
- leetcode 29 -- Divide Two Integers
- leetcode 29:Divide Two Integers
- leetcode 29: Divide Two Integers
- Leetcode#29 Divide Two Integers
- LeetCode(29)Divide Two Integers
- LeetCode-29 Divide Two Integers
- Leetcode#29||Divide Two Integers
- leetcode-29 Divide Two Integers
- Dungeon Master
- Android:学习笔记(一)
- Excel Sheet Column Number
- Unity3D汽车物理引擎的安卓重力感应适配
- Android UI之QuickContactBasge
- LeetCode(29)Divide Two Integers
- 【POJ2236】【Wireless Network】【注意初始化-1】【并查集】【连通性】
- 最小生成树+Hdu4786
- MFC自绘-WzdDialog窗口类
- 精灵可以通过setColor来设置颜色,还原为原来的颜色的办法
- Lowest Bit
- 如何让IE支持自定义协议
- Java模拟公司置办货物系统(二)
- hdu1029 Ignatius and the Princess IV