LeetCode-29-Divide Two Integers
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算法描述:
Given two integers dividend
and divisor
, divide two integers without using multiplication, division and mod operator.
Return the quotient after dividing dividend
by divisor
.
The integer division should truncate toward zero.
Example 1:
Input: dividend = 10, divisor = 3 Output: 3
Example 2:
Input: dividend = 7, divisor = -3 Output: -2
Note:
- Both dividend and divisor will be 32-bit signed integers.
- The divisor will never be 0.
- Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [?231, 231 ? 1]. For the purpose of this problem, assume that your function returns 231 ? 1 when the division result overflows.
解题思路:用加法或者向左移动实现乘法或者除法。这道题需要考虑的细节比较多。1 为避免溢出,采用长整型。2 长整型的绝对值函数为labs(long long a, long long b),3 考虑特殊溢出情况。
int divide(int dividend, int divisor) { if(!divisor || (divisor==-1 && dividend == INT_MIN)) return INT_MAX; long long ldd = labs(dividend); long long ldv = labs(divisor); int res = 0; res = divideLong(ldd, ldv); if((dividend < 0 && divisor > 0 ) || (dividend > 0 && divisor < 0)) return -(int)res; else return (int)res; } long long divideLong(long long ldd, long long ldv){ if(ldd < ldv) return 0; long long sum = ldv; long long res = 1; while((sum + sum) <= ldd){ sum += sum; res += res; } return res + divideLong(ldd-sum, ldv); }
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