LeetCode 401. Binary Watch 题解（C++）

最新推荐文章于 2022-09-11 19:37:20 发布

原创最新推荐文章于 2022-09-11 19:37:20 发布 · 1.3k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #c++

LeetCode 专栏收录该内容

53 篇文章

订阅专栏

这篇博客介绍了LeetCode 401题——二进制手表的问题，讨论了如何使用C++解决这个问题。内容包括题目描述、示例、补充信息以及两种解题思路：直接遍历和回溯法。提供了相应的代码实现。

LeetCode 401. Binary Watch 题解（C++）

题目描述

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.

举例

Input: n = 1
Return: [“1:00”, “2:00”, “4:00”, “8:00”, “0:01”, “0:02”, “0:04”, “0:08”, “0:16”, “0:32”]

补充

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”.

思路

思路1

对小时h和分钟m进行遍历，总共有12*60种可能（分别为12小时和60分钟），之后将小时h右移6位（右边6位为分钟m的二进制）转换成题目要求的10个二进制，并调用bitset的count计算二进制中1的个数，若1的个数等于num，则该时间满足题目要求，这里需要注意时间的格式。

思路2

使用回溯法，递归的出口为num=0；对10个灯做深度优先搜索，注意需要剪枝，即小时要小于12，分钟要小于60，参见代码2。

代码

代码1

class Solution
{
public:
    vector<string> readBinaryWatch(int num)
    {
        vector<string> res;
        for (int h = 0; h < 12; ++h)
        {
            for (int m = 0; m < 60; ++m)
            {
                if (bitset<10>(h << 6 | m).count() == num)
                {
                    res.push_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m));
                }
            }
        }
        return res;
    }
};

代码2

class Solution 
{
    vector<int> hour = {1, 2, 4, 8}, minute = {1, 2, 4, 8, 16, 32};
public:
    vector<string> readBinaryWatch(int num) 
    {
        vector<string> res;
        helper(res, make_pair(0, 0), num, 0);
        return res;
    }

    void helper(vector<string>& res, pair<int, int> time, int num, int start_point)
    {
        if (num == 0)
        {
            if (time.second < 10)
                res.push_back(to_string(time.first) + ":0" + to_string(time.second));
            else
                res.push_back(to_string(time.first) + ":" + to_string(time.second));
            return;
        }
        for (int i = start_point; i < hour.size() + minute.size(); i ++)
        {
            if (i < hour.size())
            {
                time.first += hour[i];
                if (time.first < 12)    // "hour" should be less than 12.
                    helper(res, time, num - 1, i + 1);
                time.first -= hour[i];
            }
            else
            {
                time.second += minute[i - hour.size()];
                if (time.second < 60)    // "minute" should be less than 60.
                    helper(res, time, num - 1, i + 1);
                time.second -= minute[i - hour.size()];
            }
        }
    }
};