[Leetcode] divide two integers

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

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

Train of Thought 1: we can only use subtraction, minus divisor and Tots in sequence, and it definitely times out.

Idea 2: each time minus n times the divisor, the result is also added n times, so we need to increase the divisor by N times.

1. Extend int to long to prevent overflow.
2. Multi [I] stores the case where the divisor is increased by I times and then subtracted by the divisor.
3. Negative number processing.
   

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));    }}

 

 

Refer:

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

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