leetcode 405. Convert a Number to Hexadecimal
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Given an integer, write an algorithm to convert it to hexadecimal(16进制的数). For negative integer, two’s complement method(就是取反加一啦) is used.
Note:
- All letters in hexadecimal (
a-f
) must be in lowercase. - The hexadecimal string must not contain extra leading
0
s. If the number is zero, it is represented by a single zero character'0'
; otherwise, the first character in the hexadecimal string will not be the zero character. - The given number is guaranteed to fit within the range of a 32-bit signed integer.
- You must not use any method provided by the library which converts/formats the number to hex directly.
Example 1:
Input:26Output:"1a"
Example 2:
Input:-1Output:"ffffffff"
把十进制数转化为十六进制数。正数很简单,负数怎么做?
负数转换成二进制分为3步:
1、首先将负数转换为对应的原码
-3的原码为(也就是+3转换成二进制后的字符串):
0000 0000 0000 0000 0000 0000 0000 0011
2、再将原码的每一位做取反操作得到反码。 取反操作:0变为1,1变为0; 取反后的结果即为:
1111 1111 1111 1111 1111 1111 1111 1100
3、将反码+1得到补码
1111 1111 1111 1111 1111 1111 1111 1101
负数转换成十六进制,只需在补码(二进制)的基础上,4位合成一位计算:
将-3的二进制从右至左每4位合并为一个单元,即:
1111 1111 1111 1111 1111 1111 1111 1101
计算后为: FFFFFFFD
其实正数也可以这么做,如26为:0000 0000 0000 0000 0000 0000 0001 1010,就是1A(删去了前导0)。
因此,无论是正数还是负数,都可以先转化为二进制,再 每4位组合,来计算十六进制。
package leetcode;import java.util.HashMap;public class Convert_a_Number_to_Hexadecimal_405 {public String toHex(int num) {if(num==0){return "0";}HashMap<Integer, Character> map=new HashMap<Integer, Character>();map.put(0,'0');map.put(1,'1');map.put(2,'2');map.put(3,'3');map.put(4,'4');map.put(5,'5');map.put(6,'6');map.put(7,'7');map.put(8,'8');map.put(9,'9');map.put(10,'a');map.put(11,'b');map.put(12,'c');map.put(13,'d');map.put(14,'e');map.put(15,'f');int[] bits=new int[32];long absNum=Math.abs((long)num);int pointer=31;while(absNum>0){if(absNum%2==1){bits[pointer]=1;}absNum=absNum/2;pointer--;}if(num<0){reverseAddOne(bits);}String result="";boolean isLeadingZero=true;for(int i=0;i<8;i++){int one=bits[i*4];int two=bits[i*4+1];int three=bits[i*4+2];int four=bits[i*4+3];if(isLeadingZero&&one==0&&two==0&&three==0&&four==0){continue;//前导0 可以忽略}isLeadingZero=false;//之后的0就不是前导0了int thisNumber=one*8+two*4+three*2+four;result=result+map.get(thisNumber);}return result;}public void reverseAddOne(int[] bits){//取反加一for(int i=0;i<32;i++){bits[i]=1-bits[i];//1-0=1,1-1=0}int pointer=31;while(pointer>=0){if(bits[pointer]==0){bits[pointer]=1;break;}else{bits[pointer]=0;pointer--;}}}public static void main(String[] args) {// TODO Auto-generated method stubConvert_a_Number_to_Hexadecimal_405 c=new Convert_a_Number_to_Hexadecimal_405();System.out.println(c.toHex(16));System.out.println(c.toHex(-2147483648));}}大神的方法则特别简洁:
/*Basic idea: each time we take a look at the last four digits of binary verion of the input, and maps that to a hex char shift the input to the right by 4 bits, do it again until input becomes 0.*/public class Solution { char[] map = {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'}; public String toHex(int num) { if(num == 0) return "0"; String result = ""; while(num != 0){ result = map[(num & 15)] + result; num = (num >>> 4); } return result; } }
Java中的位运算符:
>>表示右移,如果该数为正,则高位补0,若为负数,则高位补1;
>>>表示无符号右移,也叫逻辑右移,即若该数为正,则高位补0,而若该数为负数,则右移后高位同样补0。
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