[leetcode]29. Divide Two Integers 两整数相除

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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.

题目

不准乘除，不准膜。

思路

那剩下的还能用加减、位运算

代码

// Divide Two Integers
// 时间复杂度O(logn)，空间复杂度O(1)
public class Solution {
public int divide(int dividend, int divisor) {
if(dividend == 0) return 0;
if (divisor == 0) return Integer.MAX_VALUE; // 当 dividend = INT_MIN，divisor = -1时，结果会溢出
if (dividend == Integer.MIN_VALUE) {
if (divisor == -1) return Integer.MAX_VALUE;
else if (divisor < 0)
return 1 + divide(dividend - divisor, divisor);
else
return - 1 + divide((dividend + divisor), divisor);
}
if(divisor == Integer.MIN_VALUE){
return dividend == divisor ? 1 : 0;
} int a = dividend > 0 ? dividend : -dividend;
int b = divisor > 0 ? divisor : -divisor; int result = 0;
while (a >= b) {
int c = b;
for (int i = 0; a >= c;) {
a -= c;
result += 1 << i;
if (c < Integer.MAX_VALUE / 2) { // prevent overflow
++i;
c <<= 1;
}
}
} return ((dividend^divisor) >> 31) != 0 ? (-result) : (result);
}
}