405. Convert a Number to Hexadecimal

题目：

Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, two’s complement method is used.

Note:

All letters in hexadecimal (a-f) must be in lowercase.
The hexadecimal string must not contain extra leading 0s. 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:
26

Output:
"1a"

Example 2:

Input:
-1

Output:
"ffffffff"

链接：https://leetcode.com/problems/convert-a-number-to-hexadecimal/#/description

3/22/2017

这道题思路很简单，但是写起来有点啰嗦。

注意：

1. StringBuilder.substring()返回的是String，不是StringBuilder！

2. 判断StringBuilder的值是否是“”：ret.toString().equals("")。需要先变为String才可以

 1 public class Solution {
 2     public String toHex(int num) {
 3         StringBuilder sb = new StringBuilder();
 4         String tmp = new String();
 5         StringBuilder ret = new StringBuilder();
 6         HashMap<String, String> lookup = new HashMap<String, String>();
 7         lookup.put("0000", "0");
 8         lookup.put("0001", "1");
 9         lookup.put("0010", "2");
10         lookup.put("0011", "3");
11         lookup.put("0100", "4");
12         lookup.put("0101", "5");
13         lookup.put("0110", "6");
14         lookup.put("0111", "7");
15         lookup.put("1000", "8");
16         lookup.put("1001", "9");
17         lookup.put("1010", "a");
18         lookup.put("1011", "b");
19         lookup.put("1100", "c");
20         lookup.put("1101", "d");
21         lookup.put("1110", "e");
22         lookup.put("1111", "f");
23 
24         int c = 1;
25         for (int i = 0; i < 32; i++) {
26             if ((num & c) == c) sb.append('1');
27             else sb.append('0');
28             c <<= 1;
29         }
30         sb.reverse();
31         boolean leading = true;
32         for (int i = 0; i < 32; i+=4) {
33             tmp = sb.substring(i, i+4);
34             if (tmp.equals("0000") && leading) continue;
35             leading = false;
36             ret.append(lookup.get(tmp));
37         }
38         if (ret.toString().equals("")) return "0";
39         return ret.toString();
40     }
41 }

看别人的解法很有收获，其实不需要借用2进制变为16进制，直接用16进制的数组map就可以了，注意这时按位与也要变成和15进行

 1 public class Solution {
 2     
 3     char[] map = {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};
 4     
 5     public String toHex(int num) {
 6         if(num == 0) return "0";
 7         String result = "";
 8         while(num != 0){
 9             result = map[(num & 15)] + result; 
10             num = (num >>> 4);
11         }
12         return result;
13     }
14     
15     
16 }

讨论：https://discuss.leetcode.com/category/531/convert-a-number-to-hexadecimal

看这道题想起来一个问题，经常有输入n的问题，有没有想过什么时候用从n减来计算，什么时候从start（0或1或其他）计算到n呢？