［抄題］：

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".

?[暴力解法]：

時間分析：n2

空間分析：

［思維問題］：

不知道和回溯法有什么關系。一看特別麻煩，果斷用暴力解法了

［一句話思路］：

用bitCount轉化為二進制數

［輸入量］：空：?正常情況：特大：特小：程序里處理到的特殊情況：異常情況（不合法不合理的輸入）：

［畫圖］：

［一刷］：

表的小時不超過12

用String.format嚴格控制字符串格式

［二刷］：

［三刷］：

［四刷］：

［五刷］：

? [五分鐘肉眼debug的結果]：

［總結］：

［復雜度］：Time complexity: O(n2) Space complexity: O(n)

［英文數據結構或算法，為什么不用別的數據結構或算法］：

麻煩

［關鍵模板化代碼］：

［其他解法］：

［Follow Up］：

［LC給出的題目變變變］：

?[代碼風格] ：

public class Solution {/** @param : the number of "1"s on a given timetable* @return: all possible time*/public List<String> readBinaryWatch(int num) {List<String> time = new ArrayList<String>();for (int h = 0; h < 12; h++) {//12 not 24for (int m = 0; m < 60; m++) {if (Integer.bitCount(h) + Integer.bitCount(m) == num) {time.add(String.format("%d:%02d", h, m));//String's strict format }}}return time;} }; View Code

?

轉載于:https://www.cnblogs.com/immiao0319/p/8532325.html

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