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LeetCode 401 Binary Watch

LeetCode 401 Binary Watch

作者: 被称为L的男人 | 来源:发表于2017-12-10 10:08 被阅读123次

    LeetCode 排列组合 题目汇总
    LeetCode 数字 题目汇总
    LeetCode 动态规划 题目分类汇总
    干货!LeetCode 题解汇总

    题目描述

    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".
    

    题目分析

    题目可以抽象为:将n个名额分配给两组数,每一组数1的位置随机,但是都有限制;求所有满足要求的集合。

    代码

    public List<String> readBinaryWatch(int num) {
        List<String> result = new ArrayList<>();
        int[] hour = {8, 4, 2, 1};
        int[] minute = {32, 16, 8, 4, 2, 1};
        for(int i = 0; i <= num; i++) {
            List<Integer> hours = gen(hour, i);
            List<Integer> minutes = gen(minute, num - i);
            for(int h : hours) {
                if(h > 11) continue;
                for(int m : minutes) {
                    if(m > 59) continue;
                    result.add(h + ":" + (m < 10 ? "0" : "") + m);
                }
            }
        }
        
        return result;
    }
    
    private List<Integer> gen(int[] nums, int count) {
        List<Integer> res = new ArrayList<>();
        robot(nums, count, 0, 0, res);
        return res;
    }
    
    private void robot(int[] nums, int count, int pos, int out, List<Integer> res) {
        if(count == 0) {
            res.add(out);
            return;
        }
        for(int i = pos; i < nums.length; i++) {
            robot(nums, count - 1, i + 1, out + nums[i], res);
        }
    }
    

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