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".
把小时和分钟的每一位放在一个数组里，然后用回溯的方法去在数组里选定好的几个数，选完几个计算出和，然后添加到一个List里，主程序拿到这个list，用for循环遍历list里面的每个数，超出范围的，也就是12和60的，就跳过去，没超出的就按要求输出。
public class Solution {
public List<String> readBinaryWatch(int num) {
List<String> res = new ArrayList<>();
int[] nums1 = new int[]{8, 4, 2, 1}, nums2 = new int[]{32, 16, 8, 4, 2, 1};
for(int i = 0; i <= num; i++) {
List<Integer> list1 = generateDigit(nums1, i);
List<Integer> list2 = generateDigit(nums2, num - i);
for(int num1: list1) {
if(num1 >= 12) continue;
for(int num2: list2) {
if(num2 >= 60) continue;
res.add(num1 + ":" + (num2 < 10 ? "0" + num2 : num2));
}
}
}
return res;
}
private List<Integer> generateDigit(int[] nums, int count) {
List<Integer> res = new ArrayList<>();
generateDigitHelper(nums, count, 0, 0, res);
return res;
}
private void generateDigitHelper(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++) {
generateDigitHelper(nums, count - 1, i + 1, sum + nums[i], res);
}
}
}