401. 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".

連接：https://leetcode.com/problems/binary-watch/#/description

3/22/2017

已經不會作題了，看來仍是得好好學算法

抄別人的算法，優勢是好簡單啊，缺點是我仍是不會backtracking...

注意Integer.bitCount()這個function頗有用啊

1 public List<String> readBinaryWatch(int num) {
2     List<String> times = new ArrayList<>();
3     for (int h=0; h<12; h++)
4         for (int m=0; m<60; m++)
5             if (Integer.bitCount(h * 64 + m) == num)
6                 times.add(String.format("%d:%02d", h, m));
7     return times;        
8 }

別人的dfs作法：由於我不會

 1 class Solution(object):
 2     def readBinaryWatch(self, n):
 3         
 4         def dfs(n, hours, mins, idx):
 5             if hours >= 12 or mins > 59: return
 6             if not n:
 7                 res.append(str(hours) + ":" + "0" * (mins < 10) + str(mins))
 8                 return
 9             for i in range(idx, 10):
10                 if i < 4: 
11                     dfs(n - 1, hours | (1 << i), mins, i + 1)
12                 else:
13                     k = i - 4
14                     dfs(n - 1, hours, mins | (1 << k), i + 1)
15         
16         res = []
17         dfs(n, 0, 0, 0)
18         return res

其餘算法連接：https://discuss.leetcode.com/category/526/binary-watch