401. Binary Watch

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".

首先需要得到h和m的list，再进行排列组合。关于得到h和m的list，可以采用递归的方法。代码如下：

public class Solution {    HashMap<Integer, List<Integer>> map;    int[] nums = new int[]{1,2,4,8,16,32};        public List<String> readBinaryWatch(int num) {        map = new HashMap<Integer, List<Integer>>();        List<String> res = new ArrayList<String>();        for (int hNum = 0; hNum <= num; hNum ++) {            if (hNum > 4) {                continue;            }            List<Integer> hList = getList(hNum);            List<Integer> mList = getList(num - hNum);            for (int h: hList) {                if (h >= 12) {                    continue;                }                for (int m : mList) {                    if (m >= 60) {                        continue;                    }                    res.add(h + ":" + (m >= 10? m: "0" + m));                }            }        }        return res;    }        private List<Integer> getList(int count) {        if (map.containsKey(count)) {            return map.get(count);        }        List<Integer> list = new ArrayList<Integer>();        getListhepler(count, 0, 0, list);        map.put(count, list);        return list;    }        private void getListhepler(int count, int start, int sum, List<Integer> list) {        if (count == 0) {            list.add(sum);            return;        }        for (int i = start; i < nums.length; i ++) {            getListhepler(count - 1, i + 1, sum + nums[i], list);        }    }}

还有一种穷举的方法，代码如下：

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;        }