【LeetCode】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”.

想吐槽一下 这个EASY题目AC的还真的不容易，花了我不少时间 - -。
主要思路是用一个0,1数组来模拟这十个LED灯，然后对这个数组进行全排列，
看有多少种符合条件的情况。

class Solution {public:    vector<string> ans;    int calTime(int *a,int l,int r){        int sum=0;        for(int i=l;i<=r;i++){            sum=sum*2+a[i];        }        return sum;    }    bool isNeedSwap(int *a,int l,int r){    //查找在[l,r)区间中，有没有和a[r]相同的数字        for(int i=l;i<r;i++){            if(a[i]==a[r]){                return false;            }        }        return true;    }    string timeToString(int hour,int min){        string s;        stringstream ss,s1;        ss<<hour;        s+=ss.str();        s+=':';        s1<<min;        if(min<10)        s+='0';        string r=s1.str();        s+=r;        return s;    }    void permutation(int *a,int l,int r){   //全排列        if(l==r){            int hour=calTime(a,0,3);            int minute=calTime(a,4,9);            if(hour<12&&minute<60){                ans.push_back(timeToString(hour,minute));            }        }else{            for(int i=l;i<=r;i++){                if(!isNeedSwap(a,l,i))continue;   //考虑有重复数字的情况                swap(a[i],a[l]);                permutation(a,l+1,r);                swap(a[i],a[l]);            }        }    }    vector<string> readBinaryWatch(int num) {        int a[10];        for(int i=0;i<10;i++){            if(i<10-num){                a[i]=0;            }else{                a[i]=1;            }        }        permutation(a,0,9);        return ans;    }};