letecode [401] - Binary Watch

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".

题目大意：

　　给定一个二进制手表，小时由：1、2、4、8组成，分钟由1、2、4、8、16、32组成。当使用n个数字时，求得所有可能表示的时间。

理  解：

　　看的别人的方法。

　　暴力法：遍历0：00-11：59,判断当前时间中小时和分钟二进制中1的个数。

代 码 C++：

class Solution {
public:
    int countOne(int n){
        int res = 0;
        while(n){
            n = n&(n-1);
            res++;
        }
        return res;
    }
    
    vector<string> readBinaryWatch(int num) {
        vector<string> res;
        string str;
        for(int i=0;i<12;++i){
            for(int j=0;j<60;++j){
                if(countOne(i)+countOne(j)==num){
                    str = j>=10?to_string(j):"0"+to_string(j);
                    str = to_string(i) + ":" + str;
                    res.push_back(str);
                }
            }
        }
        return res;
    }
};

运行结果：

　　执行用时 :4 ms, 在所有 C++ 提交中击败了94.67%的用户

　　内存消耗 :8.7 MB, 在所有 C++ 提交中击败了64.21%的用户