401. Binary Watch

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".

My Solutions:

用backtracking的思想，把所有时针分钟有可能的答案找到，再组合起来

class Solution {
    public List<String> readBinaryWatch(int num) {
        List<String> res = new ArrayList<>();
        
        int[] hours = new int[]{1,2,4,8};
        int[] mins = new int[]{1,2,4,8,16,32};
        
        if (num > 10 || num < 0) return res;
        
        for (int i = 0; i <= num; i++) {
            List<Integer> l1 = generateDigit(hours, i);
            List<Integer> l2 = generateDigit(mins, num - i);
            
            for (int i1 : l1) {
                if (i1 >= 12) continue;
                for (int i2 : l2) {
                    if (i2 >= 60) continue;
                    res.add(i1 + ":" + (i2 < 10 ? "0" + i2 : i2));
                }
            }
        }
        
        return res;
    }
    
    public List<Integer> generateDigit(int[] nums, int count) {
        List<Integer> res = new ArrayList<>();
        helper(nums, count, 0, 0, res);
        return res;
    }
    
    //count = 还剩几个点点；pos = 点点在可能的数字的第几位；sum = 所有点点加起来的数字，即时针/分针数字
    public void helper(int[] nums, int count, int pos, int sum, List<Integer> res) {
        if (count == 0) {
            res.add(sum);
            return;
        }
        
        for (int i = pos; i < nums.length; i++) {
            helper(nums, count - 1, i + 1, sum + nums[i], res);
        }
    }
}