[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;}