[LeetCode] 401. Binary Watch 二进制表

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".

用二进制数字表示的表，小时用4个二进制数表示，分钟用6个二进制数表示，亮的地方代表1，不亮的代表0。给定一个非负整数n，代表现在亮的的个数，问能表示的所有可能时间？

解法1：DFS，遍历每一种可能。

解法2：枚举小时h和分钟m，然后判断二进制1的个数是否等于num。

解法3：位运算（Bit Manipulation）10盏灯泡的燃亮情况可以通过0-1024进行表示，然后计数二进制1的个数即可。利用位运算将状态拆分为小时和分钟。

Java:

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public class Solution {    public List<String> readBinaryWatch(int num) {         List<String> list = new ArrayList<>();         int timecode[] = new int[10];         dfs(timecode, 0, 0, list, num);         return list;     }       private void dfs(int[] timecode, int i, int k, List<String> list, int num) {         if(k == num) {             String res = decodeToTime(timecode);             if(res != null)                 list.add(res);             return;         }         if(i == timecode.length) return;         timecode[i] = 1;         dfs(timecode, i+1, k+1, list, num);         timecode[i] = 0;         dfs(timecode, i+1, k, list, num);     }       //输出时间，即输出可能的时间，要是时间不对则输出null     private String decodeToTime(int[] timecode) {         int hours = 0;         //按照位数转换时间         for(int i = 0; i < 4; i++) {             if(timecode[i] == 1) {                 hours = hours + (int)Math.pow(2, i);             }         }         int minutes = 0;         for(int i = 4; i < 10; i++) {             if(timecode[i] == 1) {                 minutes = minutes + (int)Math.pow(2, i-4);             }         }         String min = "" + minutes;         if(minutes < 10)             min = "0" + min;         //判断时间的可行性         if(hours  >= 12  ||  minutes  >=  60)             return null;         return hours + ":" + min;     }   }

Python:

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class Solution(object):     def readBinaryWatch(self, num):         """         :type num: int         :rtype: List[str]         """         ans = []         for h in range(12):             for m in range(60):                 if (bin(h)+ bin(m)).count('1') == num:                     ans.append('%d:%02d' % (h, m))         return ans

Python:

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class Solution(object):     def readBinaryWatch(self, num):         """         :type num: int         :rtype: List[str]         """         ans = []         for x in range(1024):             if bin(x).count('1') == num:                 h, m = x >> 6, x & 0x3F                 if h < 12 and m < 60:                     ans.append('%d:%02d' % (h, m))         return ans