Leetcode: Binary Watch-白红宇

Leetcode: Binary Watch

阅读量：6817 次

发布时间：2019-06-26

本文共 2505 字，大约阅读时间需要 8 分钟。

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".

Solution 1: Bit Manipulation

use Integer.bitCount()

1 public class Solution { 2     public List
     
       readBinaryWatch(int num) { 3         List
      
        res = new ArrayList
       
        (); 4         for (int i=0; i<12; i++) { 5             for (int j=0; j<60; j++) { 6                 if (Integer.bitCount(i) + Integer.bitCount(j) == num) { 7                     String str1 = Integer.toString(i); 8                     String str2 = Integer.toString(j); 9                     res.add(str1 + ":" + (j<10? "0"+str2 : str2));10                 }11             }12         }13         return res;14     }15 }

Solution 2: Backtracking, 非常精妙之处在于用了两个数组来帮助generate digit（例如：1011 -> 11）

1 public class Solution { 2     public List
     
       readBinaryWatch(int num) { 3         int[] nums1 = new int[]{8, 4, 2, 1}, nums2 = new int[]{32, 16, 8, 4, 2, 1}; 4         List
      
        res = new ArrayList
       
        (); 5         for (int i=0; i<=num; i++) { 6             List
        
          hours = getTime(nums1, i, 12); 7             List
         
           minutes = getTime(nums2, num-i, 60); 8             for (int hour : hours) { 9                 for (int minute : minutes) {10                     res.add(hour + ":" + (minute<10? "0"+minute : minute));11                 }12             }13         }14         return res;15     }16     17     public List
          
            getTime(int[] nums, int count, int limit) {18 List
           
             res = new ArrayList
            
             ();19 getTimeHelper(res, count, 0, 0, nums, limit);20 return res;21 }22 23 public void getTimeHelper(List
             
               res, int count, int pos, int sum, int[] nums, int limit) {24 if (count == 0) {25 if (sum < limit) 26 res.add(sum);27 return;28 }29 for (int i=pos; i