BinaryWatchOPT [source code]

public class BinaryWatchOPT {
    public List<String> readBinaryWatch(int num) {
        List<String> times = new ArrayList<>();
        for (int h = 0; h < 12; h++)
            for (int m = 0; m < 60; m++)
                if (Integer.bitCount(h * 64 + m) == num)
                    times.add(String.format("%d:%02d", h, m));
        return times;        
    }
}

这个方法其实是还可以的. 只可惜当时没有想到. 虽然速度不是很好, 29(37), 但是最起码能够想到这个办法还是很好的, 比什么也没有好.

这个严格来说算是一个 search 问题? 在 domain 有限的情况下, 这个属于典型的generate and test的做法了, 应该掌握.

Problem Description

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".
Difficulty:Easy
Category:Algorithms
Acceptance:44.81%
Contributor: LeetCode
Companies
google
Related Topics
backtracking bit manipulation
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