[Leetcode] 401. Binary Watch 解题报告

题目：

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".

思路：

也是一道典型的深度优先搜索题目。在DFS中需要完成的任务是：计算有k个灯亮的时候，有多少种二进制数字的可能性（当然由于小时和分钟的取值范围不同，在DFS中需要设置不同的终止条件）。然后我们就分别遍历在num个亮的灯中，有i个处于小时位置，有num - i个处于分钟位置时的二进制可能性，并且将它们进行组合。

代码：

class Solution {public:    vector<string> readBinaryWatch(int num) {        vector<string> ans;        for(int i = max(0, num - 6); i <= min(4, num); ++i) {            vector<int> vec1, vec2;            DFS(4, i, 0, 0, vec1);          // i LEDs representing the hour are on            DFS(6, num - i, 0, 0, vec2);    // num - i LEDs representing the minite are on            for(auto val1 : vec1) {                for(auto val2 : vec2) {                    string minute = (to_string(val2).size() == 1 ? "0" : "") + to_string(val2);                    ans.push_back(to_string(val1) + ":" + minute);                }            }        }        return ans;    }private:    void DFS(int len, int k, int index, int val, vector<int>& vec) {        if(k == 0 && len == 4 && val < 12) {    // hour termination condition            vec.push_back(val);        }        if(k == 0 && len == 6 && val < 60) {    // minute termination condition            vec.push_back(val);        }        if(index == len || k == 0) {            return;        }        DFS(len, k, index + 1, val, vec);                        // case that the index-th LED is off        DFS(len, k - 1, index + 1, val + pow(2, index), vec);    // case that the index-th LED is on    }};