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Description
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”.
分析一
题目的意思是：
给你一个非负整数，代表手表的亮的灯的数目，返回所有可能的时间点的组合。
- 这道题还是easy题目，但是我感觉不容易做不出来，还是backtrack方法，暴力出所有可能的组合。代码看似复杂，其实很简单。
代码一
class Solution {
public:
vector<string> readBinaryWatch(int num) {
vector<string> res;
vector<int> hour{8,4,2,1},minute{32,16,8,4,2,1};
for(int i=0;i<=num;i++){
vector<int> hours=generate(hour,i);
vector<int> minutes=generate(minute,num-i);
for(int h:hours){
if(h>11) continue;
for(int m:minutes){
if(m>59) continue;
res.push_back(to_string(h)+(m<10 ? ":0":":")+to_string(m));
}
}
}
return res;
}
vector<int> generate(vector<int> nums,int cnt){
vector<int> res;
solve(res,cnt,0,0,nums);
return res;
}
void solve(vector<int>& res,int cnt,int start,int out,vector<int> nums){
if(cnt==0){
res.push_back(out);
return ;
}
for(int i=start;i<nums.size();i++){
solve(res,cnt-1,i+1,out+nums[i],nums);
}
}
};