leetcode 401. Binary Watch

题目简述：
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).
Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.

算法思路：
设 i 个小时的bit亮，n-i个分钟的bit亮，分别算出i个bit的若干种组合和n-i个bit的若干种组合，将这2个组合得到的结果合并就可以得到时刻了，i 从0到n。

提交结果：
Runtime: 2 ms, faster than 79.31% of Java online submissions for Binary Watch.

参考代码：

public static List<String> readBinaryWatch(int num) {
        List<String> result = new ArrayList<String>();
        int[] hour = {1,2,4,8};
        int[] min = {1,2,4,8,16,32};
        for(int i=0;i<=num;i++){
			List<String> resHour = new ArrayList<String>();
			List<String> resMin = new ArrayList<String>();
			List<Integer> temp = new ArrayList<Integer>();
			//找到小时的组合
			findHour(hour, resHour, 0, i, temp);
			temp.clear();
			//找到分钟的组合
			findMin(min, resMin, 0, num - i, temp);
			// 时与分组合成时刻
			for (int j = 0; j < resHour.size(); j++) {
				for (int k = 0; k < resMin.size(); k++) {
					result.add(resHour.get(j) + ":" + resMin.get(k));
				}
			}
        }
        return result;
    }
    public static void findHour(int[] hour,List<String> result,int index,int count,List<Integer> temp){
    	if(count==temp.size()){
    		int sum=0;
    		for(Integer key:temp)
    			sum += key;
    		//最大11时
    		if(sum<=11)
    			result.add(String.valueOf(sum));
    		return; 		
    	}
    	for(int i=index;i<hour.length;i++){
    		temp.add(hour[i]);
    		findHour(hour, result, i+1, count, temp);
    		//恢复 findHour()之前的样子，尝试找下一个元素
    		temp.remove(temp.size()-1);
    	}
    }
    public static void findMin(int[] min,List<String> result,int index,int count,List<Integer> temp){
    	if(count==temp.size()){
    		int sum=0;
    		for(Integer key:temp)
    			sum += key;
    		//最大59分钟
    		if(sum<=59){
    			if(sum<=9)   				
    				result.add('0'+String.valueOf(sum));
    			else
    				result.add(String.valueOf(sum));
    		}
    		return; 		
    	}
    	for(int i=index;i<min.length;i++){
    		temp.add(min[i]);
    		findMin(min, result, i+1, count, temp);
    		temp.remove(temp.size()-1);
    	}
    }