401. Binary Watch -Easy

Question

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.

Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.

the above binary watch reads “3:25”.

一个二进制表有两排灯，上面一排的4盏代表0-11小时，下面一排的6盏代表0-59分钟。每个LED灯。每个LED灯的表现形式为1或0，最右侧代表最低位。现在给出一个非负整数n，它代表当前亮着的LED灯的个数，返回所有可能的表的时间。

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”]

Solution

• 回溯解。我们只需要遍历n盏亮着的灯的所有组合，并且将小时 H < 12 且 分钟 M < 60 的组合保存下来即可。

  public class Solution {    public List<String> readBinaryWatch(int num) {        List<String> res = new ArrayList<>();        backtracking(0, num, new boolean[10], res);        return res;    }    /**     * 思路：利用回溯遍历所有组合，只取小于H < 12小时且M < 59分钟的组合添加到res返回     * @param start：从第以start为索引的灯开始亮     * @param num：还需要亮的灯数     * @param choose：表示10盏灯的亮暗情况（true代表亮，false代表暗）     * @param res：返回列表结果     */    public void backtracking(int start, int num, boolean[] choose, List<String> res){        // 10盏灯的代表数值（前4个代表小时，后6个代表分钟）        int[] selectable = new int[]{1, 2, 4, 8, 1, 2, 4, 8, 16, 32};        int hh = 0, mm = 0;        // 如果已经点亮完所需的盏数，那么统计当前的H和M        if(num == 0){            for(int i = 0; i < 10; i++){                if(choose[i]){                    if(i < 4) hh += selectable[i];                    else mm += selectable[i];                }            }            // 只取 H < 12 且 M < 60 的组合            if(hh < 12 && mm < 60){                if(mm < 10) res.add(hh + ":0" + mm);                else res.add(hh + ":" + mm);            }        }        // 每次从start:10中选择一盏灯点亮        for(int i = start; i < 10; i++){            choose[i] = true;            backtracking(i + 1, num - 1, choose, res);            choose[i] = false;        }    }}