[LintCode] Divide Two Integers 两数相除

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

If it is overflow, return 2147483647

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Example

Given dividend = 100 and divisor = 9, return 11.

LeetCode上的原题，请参见我之前的博客Divide Two Integers。

解法一：

class Solution {
public:
    /**
     * @param dividend the dividend
     * @param divisor the divisor
     * @return the result
     */
    int divide(int dividend, int divisor) {
        if (divisor ==  || (dividend == INT_MIN && divisor == -)) return INT_MAX;
        long long m = abs((long long)dividend), n = abs((long long)divisor), res = ;
        int sign = ((dividend < ) ^ (divisor < )) ? - : ;
        if (n == ) return sign ==  ? m : -m;
        while (m >= n) {
            long long t = n, p = ;
            while (m >= (t << )) {
                t <<= ;
                p <<= ;
            }
            res += p;
            m -= t;
        }
        return sign ==  ? res : -res;
    }
};

解法二：

class Solution {
public:
    /**
     * @param dividend the dividend
     * @param divisor the divisor
     * @return the result
     */
    int divide(int dividend, int divisor) {
        long long m = abs((long long)dividend), n = abs((long long)divisor), res = ;
        if (m < n) return ;
        while (m >= n) {
            long long t = n, p = ;
            while (m > (t << )) {
                t <<= ;
                p <<= ;
            }
            res += p;
            m -= t;
        }
        if ((dividend < ) ^ (divisor < )) res = -res;
        return res > INT_MAX ? INT_MAX : res;
    }
};

解法三：

class Solution {
public:
    /**
     * @param dividend the dividend
     * @param divisor the divisor
     * @return the result
     */
    int divide(int dividend, int divisor) {
        long long m = abs((long long)dividend), n = abs((long long)divisor), res = ;
        if (m < n) return ;
        long long t = n, p = ;
        while (m > (t << )) {
            t <<= ;
            p <<= ;
        }
        res += p + divide(m - t, n);
        if ((dividend < ) ^ (divisor < )) res = -res;
        return res > INT_MAX ? INT_MAX : res;
    }
};