[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:

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:

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:

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