029两数相除

#pragma once
#include "000库函数.h"
/******************自解***********************/
//虽然很快通过，但不满足题中不允许使用除法的要求   36ms
class Solution {
public:
    int divide(int dividend, int divisor) {
        if (divisor == 0)return 0;
        double i = (double)dividend / (double)divisor;
        if ((i < -1 * pow(2, 31)) || (i > pow(2, 31) - 1))return pow(2, 31) - 1;
        else return (int)i;
    }
};
//超出时长
class Solution {
public:
    int divide(int dividend, int divisor) {
        if (divisor == 0)return 0;
        int flag = 1;
        double k = (double)dividend;
        if (k < 0) {
            k *= -1;//防止此处溢出
            flag *= -1;
        }
        double t = (double)divisor;
        if (t < 0) {
            t *= -1;
            flag *= -1;
        }
        double i= 0;
        double n = t;
        while (k >= t+t) {
            i += 2;
            t += n;
            k -= n;

        }
        if (k >= t)
            ++i;
        i *= flag;
        if ((i < -1 * pow(2, 31)) || (i > pow(2, 31) - 1))return pow(2, 31) - 1;
        else return (int)i;
    }
};
/*********************博客答案***************************/
//采用移位运算  36ms
class Solution {
public:
    int divide(int dividend, int divisor) {
        if (divisor == 0 || (dividend == INT_MIN && divisor == -1)) return INT_MAX;
        long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
        int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
        if (n == 1) return sign == 1 ? m : -m;
        while (m >= n) {
            long long t = n, p = 1;
            while (m >= (t << 1)) {
                t <<= 1;
                p <<= 1;
            }
            res += p;
            m -= t;
        }
        return sign == 1 ? res : -res;
    }
};

class Solution {
public:
    int divide(int dividend, int divisor) {
        long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
        if (m < n) return 0;
        while (m >= n) {
            long long t = n, p = 1;
            while (m > (t << 1)) {
                t <<= 1;
                p <<= 1;
            }
            res += p;
            m -= t;
        }
        if ((dividend < 0) ^ (divisor < 0)) res = -res;
        return res > INT_MAX ? INT_MAX : res;
    }
};
//使用递归  36ms
class Solution {
public:
    int divide(int dividend, int divisor) {
        long long res = 0;
        long long m = abs((long long)dividend), n = abs((long long)divisor);
        if (m < n) return 0;
        long long t = n, p = 1;
        while (m > (t << 1)) {
            t <<= 1;
            p <<= 1;
        }
        res += p + divide(m - t, n);
        if ((dividend < 0) ^ (divisor < 0)) res = -res;
        return res > INT_MAX ? INT_MAX : res;
    }
};


void T029() {
    Solution s;
    cout << s.divide(10, 3) << endl;
    cout << s.divide(7, -3) << endl;
}