Divide Two Integers
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1.Problem
Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT.
2.Pre-Thought
题目要求不能使用乘法、除法和求余运算符,只能使用+、-运算符。
除法可理解为被除数是由几个除数相加得到的,比如8 / 2 = 4,可写成8 = 2 + 2 + 2 + 2; 7 / 2 = 2 + 2 + 2 + 1,商为3.
那么可以用减法不断地用除数去减被除数,一直到被除数小于除数为止。
3.Example
3.1普通模式: dividend = 70, divisor = 2
1) 70 - 2 = 68 -----> factor = 1,
minus = 2,
res = 0 + factor = 1
2) 68 - (2 + 2) = 66 ----->factor = factor + factor= 2,
minus = minus + minus = 4,
res = res + factor = 3
3) 66 - (4 + 4) = 54 -----> factor = factor + factor = 4,
minus = minus + minus = 8,
res = res + factor =7
... ...
3.2特殊情形1: divisor = 0
3.3特殊情形2: divisor = 1
3.3特殊情形3: divisor = -1
(普及知识: "INT_MIN"和"INT_MAX"是在头文件<limits.h>中定义的,在32位操作系统中,INT_MAX为2^32-1=2147483647.INT_MIN为-2^32=-2147483648.)按照数学知识,(-2^32)/(-1)应该为2^32,但是2^32这个数超出了正数的表示范围.)
3.4特殊情形4: dividend < 0, divisor < 0
4.Algorithm
1)对dividend和divisor均取绝对值,一方便进行正整数相减的工作;
3)对普通模式进行处理时需要注意的两点是:
5.Coding
#include<iostream>#include<cmath> // import abs#include<limits.h>using namespace std;class Solution{public: int divide(int dividend, int divisor) { int MAX_INT = INT_MAX; // 如果除数为0,那么就输出MAX_INT if(divisor == 0) return MAX_INT; // 如果除数为1,那么就输出dividend if(divisor == 1) return dividend; // 如果除数为-1,那么就输出-dividend if(divisor == -1) { if(dividend == INT_MIN) return MAX_INT; return 0 - dividend; } // 求取除数和被除数中负数的个数 int minus = 0; if(dividend < 0) minus += 1; if(divisor < 0) minus += 1; dividend = abs(dividend); divisor = abs(divisor); int res = 0; // 处理INT_MIN的除法(divisor=INT_MIN),因为abs(INT_MIN)=INT_MIN if(divisor == INT_MIN) { if(dividend == INT_MIN) return 1; else return 0; } bool flag_min = false; if(dividend == INT_MIN) { flag_min = true; dividend = INT_MAX; } // 如果被除数小于除数,则输出0 if(dividend < divisor) return 0; // 如果被除数不小于除数,则做减法操作 while(dividend >= divisor) { int factor = 1; res += factor; int minus_current = divisor; dividend -= minus_current; if(flag_min) { flag_min = false; dividend += 1; } while(true) { int minus_temp = minus_current; minus_current += minus_current; if(minus_temp >= 0 && minus_current < 0) break; int temp = dividend - minus_current; if(temp < 0) break; else { dividend = dividend - minus_current; if(flag_min) { flag_min = false; dividend += 1; } res = res + factor + factor; factor = factor + factor; } }; } // 如果商应该是负数,那么就对结果取相反数 bool flag = minus & 1; if(flag) res = 0 - res; return res; }};int main(){ int a = INT_MIN; int b = a + 1; int res = 0; Solution sol; res = sol.divide(a, b); cout<<res<<endl; return 0;}
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