[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。

1. int扩展成long防止溢出。
2. multi[i]存储除数扩大i倍的情况，然后用被除数减。
3. 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：

http://blog.csdn.net/doc_sgl/article/details/12841741

http://blog.csdn.net/linhuanmars/article/details/20024907