Binary Watch

https://leetcode.com/problems/binary-watch/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".

Thoughts

一种简便的方法是遍历所有可能的小时和分钟, 看有多少个bits.

还有就是搜索, 依次把所有可能的小时和分钟的组合写出来.

Code

class Solution {
    private void dfs(int limit, int remains, int pos, boolean[] num, Set<Integer> res) {
        if (remains == 0) {
            StringBuilder sb = new StringBuilder();
            for (boolean b : num) {
                sb.append(b ? '1' : '0');
            }
            res.add(Integer.parseInt(sb.toString(), 2));
            return;
        }

        for (int i = pos; i < limit; i++) {
            num[i] = true;
            StringBuilder sb = new StringBuilder();
            for (boolean b : num) {
                sb.append(b ? '1' : '0');
            }
            dfs(limit, remains - 1, i + 1, num, res);
            num[i] = false;
        }
    }

    public List<String> readBinaryWatch(int num) {
        List<String> res = new ArrayList<>();
        for (int i = 0; i <= 4 && i <= num; i++) {
            Set<Integer> hours = new HashSet<>();
            System.out.println(i + "h");
            dfs(4, i, 0, new boolean[4], hours);
            System.out.println((num - i) + "m");
            Set<Integer> mins = new HashSet<>();
            dfs(6, num - i, 0, new boolean[6], mins);
            for (int hour : hours) {
                if (hour >= 12) {
                    continue;
                }
                for (int min : mins) {
                    if (min >= 60) {
                        continue;
                    }
                    res.add(hour + ":" + (min < 10 ? "0" : "") + min);
                }
            }
        }

        return res;
    }
}

Analysis

时间复杂度O(2 ^ 4 + 2 ^ 6).