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