【leetcode】【easy】401. Binary Watch​​​​​​​

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".

题目链接：https://leetcode-cn.com/problems/binary-watch/

 

思路

总体思路还是回溯法。

只是对于数字代表的含义需要进行判定。

最后生成的值需要判断其是否符合时间的要求。

class Solution {
public:
    vector<string> res;
    vector<string> readBinaryWatch(int num) {
        if(num<0 || num>8) return res;
        if(num==0){
            res.push_back("0:00");
        }else{
            getTime(num, 0, 0, 0);
        }
        return res;
    }
    void getTime(int num, int hour, int min, int idx){
        if(num==0){
            if(hour<=11 && min<=59){
                string time = to_string(hour) + ":" + (min<10?"0":"") + to_string(min);
                res.push_back(time);
            }
            return;
        }
        for(int i=idx; i<=10-num; ++i){
            int nhour = hour, nmin = min;
            if(i/6){
                nhour += pow(2, i%6);
            }else{
                nmin += pow(2, i);
            }
            getTime(num-1, nhour, nmin, i+1);
        }
        return;
    }
};

 

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来源： https://blog.csdn.net/lemonade13/article/details/104550653