题目
将两个整数相除，要求不使用乘法、除法和 mod 运算符。
如果溢出，返回 2147483647 。
样例 
Java Code Java Code Java Code Java Code
给定被除数 = 100 ，除数 = 9，返回 11
解题
15%的通过率，减法，位运算？表示不知道如何下手。
法一：利用减法，超时，人工直接去除的一些情况太流氓。
public class Solution { /** * @param dividend the dividend * @param divisor the divisor * @return the result */ public int divide(int dividend, int divisor) { // Write your code here if(dividend == 0) return 0; if(dividend == divisor) return 1; int count = 0; int flag1 = 1; int flag2 = 1; if(dividend == Integer.MIN_VALUE && divisor ==1) return dividend; if(dividend == Integer.MIN_VALUE && divisor == -1) return 2147483647; if(dividend<0){ flag1 = -1; dividend = -dividend; } if(divisor<0){ flag2 = -1; divisor = -divisor; } // dividend = 2147483647; // divisor = 2; if(divisor == 1) return dividend*flag1*flag2; while(dividend >= divisor){ dividend -= divisor; count +=1; } return count*flag1*flag2; }} 法二：批量减法，，但是还是超市，除数是1的时候超时，是1的时候我单独处理，是2的时候超时
public class Solution { /** * @param dividend the dividend * @param divisor the divisor * @return the result */ public int divide(int dividend, int divisor) { // Write your code here if(dividend == 0) return 0; if(dividend == divisor) return 1; int count = 0; int flag1 = 1; int flag2 = 1; if(dividend == Integer.MIN_VALUE && divisor ==1) return dividend; if(dividend == Integer.MIN_VALUE && divisor == -1) return 2147483647; if(dividend<0){ flag1 = -1; dividend = -dividend; } if(divisor<0){ flag2 = -1; divisor = -divisor; } // dividend = 2147483647; // divisor = 2; if(divisor == 1) return dividend*flag1*flag2; while(dividend >= divisor){ int sum = divisor; int count1 = 1; while(sum + sum <= dividend){ count1 += count1; sum += sum; } dividend -= sum; count += count1; } return count*flag1*flag2; }} 法三：利用位运算
，感觉与法二很像的，我把二中减肥换成位运算，也是运行超时，我不理解，直接用他的程序就可以通过。
public class Solution { /** * @param dividend the dividend * @param divisor the divisor * @return the result */ public int divide(int dividend, int divisor) { // Write your code here if(divisor ==0) return Integer.MAX_VALUE; if(divisor == -1 && dividend == Integer.MIN_VALUE) return Integer.MAX_VALUE; int count = 0; long pDividend = Math.abs((long)dividend); long pDivisor = Math.abs((long)divisor); while(pDividend >= pDivisor){ int count1 = 0; while((pDivisor< <= pDividend){ count1++; } count += 1<<(count1 - 1); pDividend -=(pDivisor<<(count1 - 1)); } if( dividend >0 && divisor >0 || dividend<0 && divisor<0) return count; else return -count; }} 博客中的位运算改成减法的也可以通过，就是我自己搞的减法的就是能通过，无法理解
public class Solution { /** * @param dividend the dividend * @param divisor the divisor * @return the result */ public int divide(int dividend, int divisor) { // Write your code here if(divisor ==0) return Integer.MAX_VALUE; if(divisor == -1 && dividend == Integer.MIN_VALUE) return Integer.MAX_VALUE; int count = 0; long pDividend = Math.abs((long)dividend); long pDivisor = Math.abs((long)divisor); while(pDividend >= pDivisor){ int count1 = 1; long sum = pDivisor; while(( sum + sum)<= pDividend){ count1 += count1; sum +=sum; } count += count1; pDividend -= sum; } if( dividend >0 && divisor >0 || dividend<0 && divisor<0) return count; else return -count; }}