29. Divide Two Integers

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1.问题描述

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

If it is overflow, return MAX_INT.

2.解题思路

采用二分法，每次让除数以2倍的形式递增，只要递增结果小于被除数，则继续递增，如果递增时超过了，则记住递增前的数，然后用被除数减去递增前的数，再次倍数递增，直到被除数的值小于除数

3.代码实现

class Solution {public:long long myAbs(const int x) {if (x > 0)return x;return -(static_cast<long long>(x));}int divide(int dividend, int divisor) {if (divisor == 0||dividend == 0)return 0;long long count = 0;long long tmp = 0;long long absDividend = myAbs(dividend);long long absDivisor = myAbs(divisor);long long sum = absDivisor;while (absDividend >= absDivisor) {tmp = 1;sum = absDivisor;while (sum + sum <= absDividend) {sum += sum;tmp += tmp;}count += tmp;absDividend -= sum;}if (divisor < 0 && dividend>0 || divisor > 0 && dividend < 0)count =  -count;if (count > 2147483647)return 2147483647;return count;}};

4.结果截图

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