0401. Binary Watch (E)

题目

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”.

题意

一个特立独行的手表，用二进制位来表示时间。4位表示小时，6位表示分钟。求在10位中选出n位一共可以得到的时间组合。

思路

回溯法。需要注意其中的坑：小时最大值只能为11，分钟最大值只能为59。

另一种暴力法是，小时数只有12种可能，分钟数只有60种可能，直接暴力搜索每一种组合，判断它们二进制1的位数相加是否为n。

代码实现

Java

回溯法

class Solution {
    public List<String> readBinaryWatch(int num) {
        List<String> ans = new ArrayList<>();
        int[] available = { 1, 2, 4, 8, 101, 102, 104, 108, 116, 132 };
        dfs(num, new int[2], available, 0, ans);
        return ans;
    }
    private void dfs(int num, int[] time, int[] available, int index, List<String> ans) {
        if (num == 0) {
            ans.add(time[0] + ":" + (time[1] < 10 ? "0" + time[1] : time[1]));
            return;
        }
        if (index == 10) {
            return;
        }
        if (available[index] > 100 && time[1] + available[index] - 100 < 60) {
            time[1] += available[index] - 100;
            dfs(num - 1, time, available, index + 1, ans);
            time[1] -= available[index] - 100;
        } else if (available[index] < 100 && time[0] + available[index] < 12) {
            time[0] += available[index];
            dfs(num - 1, time, available, index + 1, ans);
            time[0] -= available[index];
        }
        dfs(num, time, available, index + 1, ans);
    }
}

暴力法

class Solution {
    public List<String> readBinaryWatch(int num) {
        List<String> ans = new ArrayList<>();
        for (int h = 0; h < 12; h++) {
            for (int m = 0; m < 60; m++) {
                if (countOne(h) + countOne(m) == num) {
                    ans.add(h + ":" + (m < 10 ? "0" + m : m));
                }
            }
        }
        return ans;
    }
    private int countOne(int num) {
        int count = 0;
        while (num != 0) {
            if ((num & 1) == 1) {
                count++;
            }
            num >>= 1;
        }
        return count;
    }
}