Divide two integers

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尼玛考虑负数的情况！而且还有溢出的情况。。。。 -（-2147483648）溢出啦亲。。。

bit operator的妙用，左移实现乘2功能

public class Solution {    public int divide(int dividend, int divisor) {        long a = Math.abs((long)dividend);        long b = Math.abs((long)divisor);        boolean pos = false;                if (dividend > 0 && divisor > 0             || dividend < 0 && divisor < 0) {            pos = true;        }                 int ans = 0;        while (a >= b) {            int shift = 0;            while ( (b << shift) <= a) {                shift++;            }            ans += (1 << (shift - 1));            a -= (b << (shift - 1));        }        return pos? ans : -ans;    }}

如果允许乘法，试着用binary search做了下

//注意dividend是0的情况，这个时候sum总是比被除数大的！//尼玛考虑负数的情况！而且还有溢出的情况。。。。 -（-2147483648）溢出啦 亲。。。//要从O(n) -> O(lgn) -> O(1)public class Solution {    public int divide(int dividend, int divisor) {        long a = Math.abs((long)dividend);        long b = Math.abs((long)divisor);        boolean pos = false;                if (dividend > 0 && divisor > 0             || dividend < 0 && divisor < 0) {            pos = true;        }                 int ans = 0;        long beg = 1;        long end = a;        while (beg <= end) {            long mid = (beg + end)/2;            if (b * mid <= a && b * (mid + 1) > a) {                ans = (int)mid;                break;            }            if (b * (mid + 1) <= a) {                beg = mid + 1;            } else {                end = mid - 1;            }        }        return pos? ans : -ans;    }}

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