LeetCode-401. Binary Watch (Java)

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".

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

题意

给定一个灯的数量n，然后让列出所有可能的时间

思路

感觉不像是一到简单题。多数博客提供的方法都利用回溯法以及DFS方法。

回溯法的特点

1. 递归

2. 有树形结构，对问题进行遍历

3. 碰到不符合条件的分支停止递归，进行回溯

4. 得到最终结果进行回溯，遍历下一种情况

那么关于本题

问题 ：在0-5这6个位置上点亮n盏灯。

操作 ：在位置i上点亮一盏灯。

操作后的子问题 ：在i+1到5这些位置上可以点亮n-1盏灯。

递归结束条件 ：num==0,即灯全部亮了，则得到一个解，递归终止，逐步返回。

代码

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);            }    }}

其中关键的代码是generateDigitHelper方法。为了方便理解，在此做一个假设，然后进行图解。

对于需要遍历各种情况的题目，可以运用此方法。