[leetcode] Divide Two Integers

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

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

思路1：看樣只能用減法了，依次減去除數，TOTS，肯定逾時。

思路2：每次減去N倍的除數，結果也加上N次，因此我們需要將除數擴大N倍。

1. int擴充成long防止溢出。
2. multi[i]儲存除數擴大i倍的情況，然後用被除數減。
3. 負數處理。
   

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));    }}

 

 

參考：

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

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