LeetCode Divide Two Integers

Description:

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.

Solution:

感觉用英文说不清了……哎……

是这样的，既然不能用乘除法，老夫的第一个反应是将被除数依次二分，dfs一下，这样有一个问题就是无法处理二分之后的+1 -1问题

既然二分，从上到下不好处理，那干脆就自下到上，将除数扩大到最大，然后再减法就好

具体描述一下

1. 将除数扩大，二分反过来就是乘以2，不过不用乘法，就用位运算 <<=1

2. 将除数扩大到小于被除数的最大数，准备阶段结束

3. 进入循环，循环条件就是除数大于等于最开始的除数初始值（divisor）以及被除数大于0

4. 每次循环都进行判断，如果a>b，那么就表示当前tot位是可以有一个的，也就是ans+=1l << tot;（这里的'l'保证是long类型的），a-=b

5. 然后每次循环都会将位数tot-1，b除以2

<span style="font-size:18px;">public class Solution {public int divide(int dividend, int divisor) {long a = dividend;long c = divisor, b = divisor;int flag = 1;if (a < 0) {a = -a;flag = -flag;}if (b < 0) {b = -b;c = -c;flag = -flag;}long ans = 0;int tot = 0;while (a > b) {if ((b << 1) > a)break;b <<= 1;tot++;}while (a > 0) {if (b < c)break;if (a >= b) {ans += 1l << tot;a = a - b;}b >>= 1;tot--;}ans = ans * flag;if (ans > Integer.MAX_VALUE)return Integer.MAX_VALUE;if (ans < Integer.MIN_VALUE)return Integer.MAX_VALUE;return (int) ans;}public static void main(String[] args) {Solution s = new Solution();System.out.println(s.divide(-2147483648, -1));}}</span>

这里换一个添加一个更加简洁的写法

<span style="font-size:18px;">public class Solution {public int divide(int dividend, int divisor) {int sign = 1;if (dividend < 0)sign = -sign;if (divisor < 0)sign = -sign;long a = Math.abs((long) dividend);long b = Math.abs((long) divisor);int move = 0;while (b << move < a)move++;long ans = 0l;while (move >= 0) {if ((b << move) <= a) {a -= (b << move);ans = ans + (1l << move);}move--;}ans = sign * ans;if (ans > Integer.MAX_VALUE)return Integer.MAX_VALUE;if (ans < Integer.MIN_VALUE)return Integer.MAX_VALUE;return (int) ans;}}</span>