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

If it is overflow, return MAX_INT.

一刷
題解：
核心是判斷符號后使用bit shift以及減法凹炸。做題過程中要仔細處理各種corner case减响。比如:

divisor = 0
　　dividend 或者 divisor = Integer.MIN_vALUE
divisor = Integer.MIN_VALUE

• if dividend = Integer.MIN_VALUE, return 1
  return 0
  dividend = Integer.MIN_VALUE
• if divisor = -1, return Integer.MAX_VALUE
• if divisor > 0, set divisor = - divisor - calculate everything in the range of negative int
  Time Complexity - O(1), Space Complexity - O(1)。

public class Solution {
    public int divide(int dividend, int divisor) {
        if(divisor == 0) return Integer.MAX_VALUE;
        if(dividend == 0) return 0;
        boolean hasSameSign = (divisor>0) ^ (dividend<0);
        int res = 0;
        if(divisor != Integer.MIN_VALUE && dividend != Integer.MIN_VALUE){
            divisor = Math.abs(divisor);
            dividend = Math.abs(dividend);
            for(int i=0; i<32; i++){//can be extended to binarysearch
                if((dividend>>(31 - i)) >=divisor){
                    dividend -= (divisor<<(31 - i));
                    res += 1<<(31 - i);
                }
            }
        }
        else{
            if(divisor == Integer.MIN_VALUE) return dividend == Integer.MIN_VALUE? 1:0;
            if(divisor == -1) return Integer.MAX_VALUE;
            if(divisor >0) divisor = -divisor; // dividend == Integer.MIN_VALUE
            for(int i=0; i<32; i++){//can be extended to binarysearch
                if((dividend>>(31 - i)) <=divisor){
                    dividend -= (divisor<<(31 - i));
                    res += 1<<(31 - i);
                }
            }
        }
        return hasSameSign? res: -res;
    }
}

二刷
思路就是分別用divisor, 2divisor, 4divisor, 8divisor去試讽坏，然后找到一個最大的但是小于dividend的值。然后dividend減去它。重復該循環(huán)食零。注意，中間值要轉成long型整數(shù)寂屏，避免Math.abs時對于corner case: Integer.MIN_VALUE和Integer.MAX_VALUE的值出現(xiàn)錯誤贰谣。

public class Solution {
    public int divide(int dividend, int divisor) {
        if(divisor==0 || (dividend == Integer.MIN_VALUE && divisor == -1)) return Integer.MAX_VALUE;
        int sign = ((dividend<0) ^ (divisor<0))? -1:1;
        long dvd = Math.abs((long)dividend);
        long dvs = Math.abs((long)divisor);
        long res = 0;
        while(dvd>=dvs){
            long temp = dvs, multiple = 1;
            while(dvd>=(temp<<1)){
                temp <<=1;
                multiple <<= 1;
            }
            dvd -= temp;
            res += multiple;
        }
        return sign==1? (int)res : (int)-res;
    }
}