[leetcode] Divide Two Integers

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

https://oj.leetcode.com/problems/divide-two-integers/

思路1：看样只能用减法了，依次减去除数，TOTS，肯定超时。

思路2：每次减去N倍的除数，结果也加上N次，因此我们需要将除数扩大N倍。

int扩展成long防止溢出。

multi[i]存储除数扩大i倍的情况，然后用被除数减。

负数处理。

public class Solution {
public int divide(int dividend, int divisor) {
if (dividend == 0 || divisor == 1)
return dividend;
long divid = dividend;
long divis = divisor;

boolean neg = false;
int result = 0;

if (dividend < 0) {
neg = !neg;
divid = -divid;
}
if (divisor < 0) {
neg = !neg;
divis = -divis;
}

long[] multi = new long[32];

for (int i = 0; i < 32; i++)
multi[i] = divis << i;

for (int i = 31; i >= 0; i--) {
if (divid >= multi[i]) {
result += 1 << i;
divid -= multi[i];
}
}

return (neg ? -1 : 1) * result;
}

public static void main(String[] args) {
System.out.println(new Solution().divide(5, 2));
System.out.println(new Solution().divide(5, -2));
System.out.println(new Solution().divide(100, 2));
System.out.println(new Solution().divide(222222222, 2));
System.out.println(new Solution().divide(-2147483648, 2));
}
}

第二遍记录:

　　结果可能溢出，考虑-2147483648 除以 -1 的情况。

　　除数为0的情况，要抛出异常么。

第三遍记录：

public int divide(int dividend, int divisor) {
if (dividend == 0 || divisor == 1)
return dividend;
long divid = dividend;
long divis = divisor;

boolean neg = false;
int result = 0;

if (dividend < 0) {
neg = !neg;
divid = -divid;
}
if (divisor < 0) {
neg = !neg;
divis = -divis;
}

for (int i = 31; i >= 0; i--) {
long tmp = divis << i;
if (divid >= tmp) {
result += 1 << i;
divid -= tmp;
}
}

return (neg ? -1 : 1) * result;
}

参考：

/article/2750079.html

/article/1378351.html