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<String> readBinaryWatch(int num) {
List<String> 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<String> 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;
}

}