Leetcode 29. Divide Two Integers

29. Divide Two Integers

Total Accepted: 70841 Total Submissions: 448890 Difficulty: Medium

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

If it is overflow, return MAX_INT.

思路：

dividend=a₁*divisor+a₂*divisor+...+a_n*divisor（a_x为2^k）。

以下情况造成溢出：
1.除数为0

2.被除数=INT_MIN，除数为-1

代码：

代码中只使用了int类型。注意INT_MIN=-2147483648，INT_MAX=2147483647。

class Solution {
public:
    int divide(int dividend, int divisor) {
        if(divisor==0||(dividend==INT_MIN&&divisor==-1)){
            return INT_MAX;
        }
        if(divisor==1){
            return dividend;
        }
        if(dividend==INT_MIN){
            if(divisor&1){
                return divide(dividend+1,divisor);
            }
            else{
                return divide(dividend>>1,divisor>>1);
            }
        }
        if(divisor==INT_MIN){
            return 0;
        }
        //符号判断
        bool sign=(dividend<0)^(divisor<0);
        //这里要防止溢出，因此到这里要满足dividend!=INT_MIN，divisor!=INT_MIN
        if(dividend<0) dividend=-dividend;
        if(divisor<0) divisor=-divisor;
        int result=0,temp,tempd;
        while(dividend>=divisor){
            tempd=divisor;
            temp=1;
            //溢出判断
            while(INT_MAX>>1>=tempd&&dividend>=tempd){
                tempd=tempd<<1;
                temp<<=1;
            }
            if(dividend<tempd){
                tempd>>=1;
                temp>>=1;
            }
            dividend-=tempd;
            result+=temp;
        }
        return sign?-result:result;
    }
};

或者：

class Solution {
public:
    int divide(int dividend, int divisor) {
        if (!divisor || (dividend == INT_MIN && divisor == -1))
            return INT_MAX;
        int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
        long long dvd = labs(dividend);
        long long dvs = labs(divisor);
        int res = 0;
        while (dvd >= dvs) { 
            long long temp = dvs, multiple = 1;
            while (dvd >= (temp << 1)) {
                temp <<= 1;
                multiple <<= 1;
            }
            dvd -= temp;
            res += multiple;
        }
        return sign == 1 ? res : -res; 
    }
};