Leetcode 401. Binary Watch 二进制的表 解题报告

1 解题思想

这道题就是说，有一个表是用二进制表示
4位表示小时
6位表示分钟
亮起来的地方表示1，反之则是0.

现在问说，如果这个表一共亮了n个灯，问可能有多少可能的时间组合呢？

其实这道题就只能搜索，搜索和不断验证每种可能了~~
因为是easy难度，所以主要参考了讨论中的

2 原题

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”.

3 AC解

//此题主要参照discuss
public class Solution {
   public List readBinaryWatch(int num) {
        List list = new ArrayList<>();
        //用来表示时间的所有可能的取值
        int timecode[] = new int[10];
        dfs(timecode, 0, 0, list, num);
        return list;
    }

    //  dfs 遍历所有可能性
    private void dfs(int[] timecode, int i, int k, List list, int num) {
        if(k == num) {
            String res = decodeToTime(timecode);
            if(res != null)
                list.add(res);
            return;
        }
        if(i == timecode.length) return;
        timecode[i] = 1;
        dfs(timecode, i+1, k+1, list, num);
        timecode[i] = 0;
        dfs(timecode, i+1, k, list, num);
    }

    //输出时间，即输出可能的时间，要是时间不对则输出null
    private String decodeToTime(int[] timecode) {
        int hours = 0;
        //按照位数转换时间
        for(int i = 0; i < 4; i++) {
            if(timecode[i] == 1) {
                hours = hours + (int)Math.pow(2, i);
            }
        }
        int minutes = 0;
        for(int i = 4; i < 10; i++) {
            if(timecode[i] == 1) {
                minutes = minutes + (int)Math.pow(2, i-4);
            }
        }
        String min = "" + minutes;
        if(minutes < 10)
            min = "0" + min;
        //判断时间的可行性
        if(hours  >= 12  ||  minutes  >=  60)
            return null;
        return hours + ":" + min;
    }

}