LeetCode 401 Binary Watch （Java）

题意：
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”.

题目略长，但是很好理解，就是二进制显示时间，时间范围：【0：00 – 11：59】
好，废话不说，先上我的代码：

public class Solution {

    public int int2bit(int x) {
        int num = 0, temp = x;
        while(temp != 0) {
            if((temp & 0x01) == 0x01) {
                num++;
            }
            temp >>= 1;
        }
        return num;
    }

    public List<String> readBinaryWatch(int num) {
        if(num < 0 || num > 10) return null;
        List<String> watches = new ArrayList<String>();
        if(num == 0) {
            watches.add("0:00");
            return watches;
        }
        for(int i = 0; i < 12; i++) {
            for(int j = 0; j < 60; j++) {
                if(int2bit(i) + int2bit(j) == num) {
                    if(j < 10)
                        watches.add(i + ":0" + j);
                    else
                        watches.add(i + ":" + j);
                }
            }
        }
        return watches;
    }
}

解题思路：输入的总的LED亮灯数目是固定的，但是小时和分钟是不固定的，但是二者的LED亮灯数目之和既是固定的；而且！小时和分钟的范围是有限且固定的，因而采用最土最直接的方法，就是遍历一遍，如果二者的LED亮灯数目之和等于输入的数目，即符合要求，add进list！
但是，怎么将整型数转换成LED亮灯数呢？说白了，就是非负整型数转byte，再计算非零位总数，代码中int2bit()方法就是这个作用，运用位运算得到结果~每次右移一位与1作按位与运算，结果为1，则num+1，循环右移直至等于0，返回结果~