[leetcode] Divide Two Integers-优快云博客

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46539871

本文介绍了一种不使用乘、除、模运算符的方法来实现整数除法，包括溢出处理，返回最大整数值。

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From : https://leetcode.com/problems/divide-two-integers/

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

If it is overflow, return MAX_INT.

class Solution { 
public: 
    int divide(int dividend, int divisor) { 
        if(divisor == 0) return INT_MAX; 
        long long int d1 = dividend, d2 = divisor;
		long long int flag = (d1>0)^(d2>0)?-1:1, ans=0;
        d1 = abs(d1); 
        d2 = abs(d2); 
        if(d1 < d2) return 0; 
        vector<long long int> cnts; 
        vector<long long int> nums; 
        for(long long int n=1; d2<=d1; d2 <<= 1,n <<= 1) { 
            nums.push_back(d2);
            cnts.push_back(n);
        }
        for(int i=nums.size()-1; i>=0; i--) { 
             if(d1>=nums[i]) {
                 ans += cnts[i];
                 d1 -= nums[i];
             }
        }
		ans *= flag;
        return -ans==INT_MIN?INT_MAX:ans; 
    } 
};

class Solution {
public:
    int divide(int dividend, int divisor) {
        if(divisor == 0) return INT_MAX; 
        long long int d1 = dividend, d2 = divisor;
		long long int flag = (d1>0)^(d2>0)?-1:1, ans=0;
        d1 = abs(d1); 
        d2 = abs(d2);
		while(d1 >= d2) {
			for(long long int base=d2, n=1; base<=d1; base<<=1, n<<=1) {
				if(base <= d1) {
					d1 -= base;
					ans += n;
				}
			}
		}
		ans *= flag;
        return -ans==INT_MIN?INT_MAX:ans; 
    }
};