【Leetcode】401. Binary Watch

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本文探讨了LeetCode上的一道经典题目——二进制手表问题。通过递归回溯的方法，实现对不同数量的LED灯亮起时，手表可能显示的所有时间组合。文章详细介绍了算法的具体实现过程。

题目链接：https://leetcode.com/contest/5/problems/binary-watch/

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.

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”.

思路：
分别对小时、分钟两个部分进行回溯，然后将结果进行组合。

算法：

    public List<String> readBinaryWatch(int k) {
        List<String> res = new ArrayList<String>();

        for (int h = 0; h <= k; h++) {
            dspCombination(hour, h, 0, true);
            dspCombination(minu, k - h, 0, false);

            for (int i : hs) {
                for (int j : ms) {
                    if (j == 0)
                        res.add("" + i + ":00");
                    else {
                        if (j / 10 == 0)
                            res.add("" + i + ":0" + j);
                        else {
                            res.add("" + i + ":" + j);
                        }
                    }
                }
            }

            hs.clear();
            ms.clear();
        }

        return res;
    }

    int hour[] = new int[] { 1, 2, 4, 8 };
    int minu[] = new int[] { 1, 2, 4, 8, 16, 32 };
    List<Integer> list = new ArrayList<Integer>();
    List<Integer> hs = new ArrayList<Integer>();
    List<Integer> ms = new ArrayList<Integer>();

    private void dspCombination(int[] nums, int k, int level, boolean flag) {
        if (list.size() == k) {
            int sum = 0;
            for (int num : list) {
                sum += num;
            }
            if (flag) {// 当前是hour
                if (sum <= 11)
                    hs.add(sum);
            } else {
                if (sum <= 59)
                    ms.add(sum);
            }
            return;
        } else if (list.size() > k) {
            return;
        } else {
            for (int i = level; i < nums.length; i++) {
                list.add(nums[i]);
                dspCombination(nums, k, i + 1, flag);
                list.remove(list.size() - 1);
            }
        }
    }