【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;
}
};