leetcode--401. Binary Watch

Posted 2020-10-04 Shihu

1、问题描述

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".

2、边界条件：1、num比较大的情况，在前面就可以处理掉，后面就可以不用考虑了；2、num = 0的情况；3、num<0不考虑，其他也在hour的循环里面过滤掉了。

3、思路：1、先确定hour的亮灯数，for循环；2、num - hour即为min的亮灯数，校验一下合法性；3、用递归的方法，分别求出hour和min的亮灯组合；4、最后将hour和min组合起来。

递归子函数思路：1、来到一个位置，有两种选择-亮灯或者不亮灯；2、选择不亮灯，亮灯数不变，index+1，但是有限制条件--剩余位置>剩余亮灯数，否则必须选择亮灯；3、选择亮灯，亮灯数-1，index+1；4、进入下一层。

base case：亮灯数==0，1、把int[] 记录的亮灯分布，计算成时/分；2、hour不用多处理，直接转为String保存，min需要判断>9?决定是否添0。

4、代码实现

class Solution {
    public List<String> readBinaryWatch(int num) {
        List<String> results = new ArrayList<>();
        for (int hourLedOnNum = 0; hourLedOnNum < 4 && hourLedOnNum <= num; hourLedOnNum++) {
            int minLedOnNum = num - hourLedOnNum;
            if (minLedOnNum >= 6) {
                continue;
            }
            List<String> hours = new ArrayList<>();
            int[] hourLeds = {0, 0, 0, 0};
            timeCombs(hours, hourLeds, hourLedOnNum, 0);

            List<String> mins = new ArrayList<>();
            int[] minLeds = {0, 0, 0, 0, 0, 0};
            timeCombs(mins, minLeds, minLedOnNum, 0);

            results.addAll(watchCombs(hours, mins));
        }
        return results;
    }

    public void timeCombs(List<String> times, int[] leds, int ledOnNum, int index) {
        if (ledOnNum == 0 || index == leds.length) {//后面条件可以去掉
            //record reuslts
            int time = 0;
            for (int i = 0; i < leds.length; i++) {
                time += leds[i] << i;
            }
            if ((leds.length == 4 && time > 11)
                || (leds.length == 6 && time > 59)) {
                return;
            }
            if (leds.length == 6 && time < 10) {//min小于10需要加0
                times.add("0" + Integer.toString(time));
            } else {
                times.add(Integer.toString(time));//Integer.toString(int i) int i 转成 String
            }
            return;
        }
        if (ledOnNum < leds.length - index) {
            timeCombs(times, leds, ledOnNum, index + 1); //not choose
        }
        leds[index] = 1; //choose
        timeCombs(times, leds, ledOnNum - 1, index + 1);
        leds[index] = 0;
    }

    public List<String> watchCombs(List<String> hours, List<String> mins) {
        List<String> results = new ArrayList<>();
        for (int i = 0; i < hours.size(); i++) {
            for (int j = 0; j < mins.size(); j++) {
                results.add(hours.get(i) + ":" + mins.get(j));
            }
        }
        return results;
    }
}

 5、时间复杂度：； 空间复杂度：

6、主要api：

int类型转成String：String Integer.toString(int i)，i是被转换，返回值为String
Integer类型转为String: Integer i = new Integer(); String str = i.toString();
类似的还有 Character.toString(char c); Character ch = new Character(); ch.toString();

7、其他解法：
从结果上面筛选，数一下bit数

    public List<String> readBinaryWatch(int num) {
        ArrayList<String> result = new ArrayList<>();
        for (int i = 0; i < 12; i++) {
            for (int j = 0; j < 60; j++) {
                if (Integer.bitCount(i) + Integer.bitCount(j) == num) {
                    result.add(String.format("%d:%02d", i, j));
                }
            }
        }
        return result;
    }