[leetcode] 401. Binary Watch 解題報告

題目鏈接：https://leetcode.com/problems/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".

思路：基本還是一道DFS的題目，分別在小時和分鐘上做DFS，給定幾個燈亮，然後把這些亮的燈枚舉分給小時和分鐘．需要注意的是剪枝，即小時必須小於１２，分鐘小於６０．然後將小時和分鐘組合即可．還有一個需要注意的是如果分鐘只有１位數，還要補０．

代碼如下：

class Solution {
public:
    void DFS(int len, int k, int curIndex, int val, vector<int>& vec)
    {
        if(k==0 && len==4 && val < 12) vec.push_back(val);
        if(k==0 && len==6 && val < 60) vec.push_back(val);
        if(curIndex == len || k == 0) return;
        DFS(len, k, curIndex+1, val, vec);
        val += pow(2, curIndex), k--, curIndex++;
        DFS(len, k, curIndex, val, vec);
    }
    
    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), DFS(6, num-i, 0, 0, vec2);
            for(auto val1: vec1)
                for(auto val2: vec2) 
                {
                    string str = (to_string(val2).size()==1?"0":"") + to_string(val2);
                    ans.push_back(to_string(val1)+":"+ str);
                }
        }
        return ans;
    }
};