LeetCode Binary Watch

原題連接在這裏：https://leetcode.com/problems/binary-watch/

題目：

A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59).

Each LED represents a zero or one, with the least significant bit on the right.

For example, the above binary watch reads "3:25".

Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.

Example:

Input: n = 1
Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]

Note:

• The order of output does not matter.
• The hour must not contain a leading zero, for example "01:00" is not valid, it should be "1:00".
• The minute must be consist of two digits and may contain a leading zero, for example "10:2" is not valid, it should be "10:02".

題解：

上面4個表示小時，下面6個表示分鐘，把小時向左移動6位 | 分鐘就是二進制表示。二進制表示中 1 bit的個數等於num時就把這個組合加到結果中。

能夠用Number of 1 Bits 或者Integer.bitCount()求出1 bit的個數.

Time Complexity: O(1), 12 * 60 * O(1). Space: O(1).

AC Java:

 1 public class Solution {
 2     public List<String> readBinaryWatch(int num) {
 3         List<String> res = new ArrayList<String>();
 4         for(int h = 0; h<12; h++){
 5             for(int m = 0; m<60; m++){
 6                 if(Integer.bitCount(h<<6 | m) == num){
 7                     res.add(String.format("%d:%02d", h, m));
 8                 }
 9             }
10         }
11         return res;
12     }
13 }