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".

题目比较简单,就是求电子表的时间。

小时序列 = {8, 4, 2, 1}

分钟序列 = {32, 16, 8, 4, 2, 1},

如果有num个灯亮了,假设小时有i个,那分钟就有num-i个,采用回溯算法,将这些组合加起来,就是答案,程序如下所示:


class Solution {
    public List readBinaryWatch(int num) {
        int hour = 0, minute = 0;
        int[] a = new int[]{8, 4, 2, 1};
        int[] b = new int[]{32, 16, 8, 4, 2, 1};
        List resultList = new ArrayList<>();
        for (int i = 0; i <= num; ++ i){
            List hourList = new ArrayList<>();
            List minuteList = new ArrayList<>();
            backTracing(a, 0, i, 0, 11, hourList);
            backTracing(b, 0, num - i, 0, 59, minuteList);
            int l0 = hourList.size(), l1 = minuteList.size();
            for (int m = 0; m < l0; ++ m){
                for (int n = 0; n < l1; ++ n){
                    resultList.add(String.valueOf(hourList.get(m)) + ":" + String.format("%02d", minuteList.get(n)));
                }
            }
        }
        Collections.sort(resultList);
        return resultList;
    }
    
    public void backTracing(int[] a, int begin, int n, int total, int max, List list){
        if (begin > a.length){
            return;
        }
        if (total > max){
            return;
        }
        if (n == 0){
            list.add(total);
            return;
        }
        for (int i = begin; i < a.length; ++ i){
            backTracing(a, i + 1, n - 1, total + a[i], max, list);
        }
    }
}


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