letcode-两数相除

题解

设未知数: Br= 125 / 3，拆进行如下拆解：

1. Br = 125 / 3
2. Br = (29 + 96)/3
3. Br = 29/3 + (32 * 3) / 3
4. Br = 29/3 + (2 * 2 * 2 * 2 * 2 * 3) / 3
5. Br = (5 + 24) / 3 + (2 * 2 * 2 * 2 * 2 * 3) / 3
6. Br = 5/3 + (8 * 3) /3 + (2 * 2 * 2 * 2 * 2 * 3) / 3
7. Br = (2 + 3)/3 + (2 * 2 * 2 * 3)/3 + (2 * 2 * 2 * 2 * 2 * 3) / 3
8. Br = 2/3 + (1 * 3)/3 + (2 * 2 * 2 * 3)/3 + (2 * 2 * 2 * 2 * 2 * 3) / 3
9. Br = 0 + 1 + 8 + 32
10. Br = 41

看到这个表达式的推演，那么我们可以使用递归进行求解了

• 为了方便计算，把除数与被除数取为负数

• 为了判断溢出，把最小负数(int)绝对值的一半定义为静态变量：halfMax = -1073741824; 当某个数大于它时二倍必定溢出

代码

class Solution {

    /** 最小负数绝对值的一半 */
    private static final int halfMax = -1073741824;

    public int divide(int dividend, int divisor) {
        if (divisor == 1) return  dividend;
        if (divisor == -1) return dividend == Integer.MIN_VALUE ? Integer.MAX_VALUE : -dividend;
        /** 两数异号：位与大于0,位或小于0  */
        boolean diff = (dividend < 0 && divisor >0) || (dividend > 0 && divisor < 0);
        /** 取为负 */
        dividend = dividend > 0 ? -dividend : dividend;
        divisor = divisor > 0 ? -divisor : divisor;
        if (dividend > divisor) return 0;
        int result = deepDivide(dividend, divisor, divisor, 1);
        return diff ?  -result : result;
    }

    private int deepDivide(int dividend, int divisor, int curr, int result){
        if (dividend > divisor) return 0;
        int dCurr = curr + curr;
        if (dCurr < dividend || curr < halfMax) {
            int btn = dividend - curr;
            if (btn > divisor){
                return result;
            }
            /** 被除数与当前值差值递归 */
            return result + deepDivide(btn, divisor, divisor, 1);
        } else {
            result += result;
            if (dCurr == dividend){
                return result;
            } else {
                /** 被除数与当前值二倍差值递归 */
                return deepDivide(dividend, divisor, dCurr, result);
            }
        }
    }
}