[leetcode]: 401. Binary Watch

1.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.



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”]

2.分析

有两种方法

方法一：

先确定小时和分钟各有多少个1，再求和计算小时和分钟。

做排列组合求和，小时有x个1，分钟有n-x个1。然后每个1出现在哪一位做排列组合。

方法二：

反向思考。

先确定小时和分钟的数字，再判断是否满足bit为1的位数为n。

小时0-11，分钟0-59.两个for循环嵌套遍历，计算当前循环的数字bit为1的数目，如果总数刚好是n则为一个符合要求的时间。

3.代码

方法1，排列组合

vector<string> readBinaryWatch(int num,int m) {
vector<int> hour = { 1,2,4,8 };
vector<int> minute = { 1,2,4,8,16,32 };
vector<string> ans;
for (int i = 0; i <= num; i++) {
unordered_set<int> hours = generateTime(hour, i);
unordered_set<int> minutes = generateTime(minute, num - i);
for (int h : hours) {
if (h < 12) {
for (int m : minutes) {
if (m < 60)
ans.push_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m));
}
}
}
}
}
unordered_set<int> generateTime(vector<int> base, int count) {
unordered_set<int> times;
permutate(base, count, 0, 0, times);
return times;
}
void permutate(vector<int> base, int count, int pos, int sum, unordered_set<int>& times) {
if (count == 0) {
times.insert(sum);
return;
}
for (int i = pos; i < base.size(); i++)
permutate(base, count - 1, i + 1, sum + base[i], times);
}

方法2，数bit为1的位数

int bitcount(int n) {
int count = 0;
while (n) {
n = n&(n - 1);
++count;
}
return count;
}
vector<string> readBinaryWatch(int num) {
vector<string> ans;
for (int i = 0; i < 12; i++) {
for (int j = 0; j < 60; j++) {
if (bitcount(i*64 + j) == num)
ans.push_back(to_string(i) + (j < 10 ? ":0" : ":") + to_string(j));
}
}
return ans;
}