前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发，mcf171专栏。这次比赛略无语，没想到前3题都可以用暴力解。
博客链接：mcf171的博客
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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 = 1Return: ["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". 这个题目因为数字范围有限，直接粗暴枚举就能解。 Your runtime beats 77.73% of java submissions. public class Solution { public List<String> readBinaryWatch(int num) { List<String> result = new ArrayList<String>(); for(int i = 0; i <= num; i ++){ List<String> hours = constructResult(i,12); List<String> mins = constructResult(num - i, 60); for(String hour : hours) for(String min : mins){ StringBuilder sb = new StringBuilder(hour); sb.append(":"); if(min.length() == 1) sb.append("0"); sb.append(min); result.add(sb.toString()); } } return result; } private List<String> constructResult(int posNum, int limit){ List<String> record = new ArrayList<String>(); for(int i = 0; i < limit; i ++){ if(check(i,posNum)) record.add(i + ""); } return record; } private boolean check(int num, int posNum){ int count = 0; boolean flag = true; while(num !=0){ if((num & 1) == 1)count ++; num = num>>> 1; if(count > posNum){flag = false; break;} } return count == posNum && flag; } }
