码迷,mamicode.com
首页 > 其他好文 > 详细

*Divide Two Integers

时间:2016-05-08 10:15:41      阅读:165      评论:0      收藏:0      [点我收藏+]

标签:

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

If it is overflow, return MAX_INT.

 

Subscribe to see which companies asked this question

 

public int divide(int dividend, int divisor) {
    //Reduce the problem to positive long integer to make it easier.
    //Use long to avoid integer overflow cases.
    int sign = 1;
    if ((dividend > 0 && divisor < 0) || (dividend < 0 && divisor > 0))
        sign = -1;
    long ldividend = Math.abs((long) dividend);
    long ldivisor = Math.abs((long) divisor);

    //Take care the edge cases.
    if (ldivisor == 0) return Integer.MAX_VALUE;
    if ((ldividend == 0) || (ldividend < ldivisor)) return 0;

    long lans = ldivide(ldividend, ldivisor);

    int ans;
    if (lans > Integer.MAX_VALUE){ //Handle overflow.
        ans = (sign == 1)? Integer.MAX_VALUE : Integer.MIN_VALUE;
    } else {
        ans = (int) (sign * lans);
    }
    return ans;
}

private long ldivide(long ldividend, long ldivisor) {
    // Recursion exit condition
    if (ldividend < ldivisor) return 0;

    //  Find the largest multiple so that (divisor * multiple <= dividend), 
    //  whereas we are moving with stride 1, 2, 4, 8, 16...2^n for performance reason.
    //  Think this as a binary search.
    long sum = ldivisor;
    long multiple = 1;
    while ((sum+sum) <= ldividend) {
        sum += sum;
        multiple += multiple;
    }
    //Look for additional value for the multiple from the reminder (dividend - sum) recursively.
    return multiple + ldivide(ldividend - sum, ldivisor);
}

reference: https://leetcode.com/discuss/57346/clean-java-solution-with-some-comment

*Divide Two Integers

标签:

原文地址:http://www.cnblogs.com/hygeia/p/5469803.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!