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[LeetCode]Minimum Time Difference(java)

我用set去重無重複則迴圈計算差值,而discuss則考慮用分鐘hash

貼一個我的答案

public class Solution {
    public int findMinDifference(List<String> timePoints) {
        int result = Integer.MAX_VALUE;
        Set<String> set = new HashSet<String>(timePoints);
        if(set.size()<timePoints.size())
            return 0;
        for(int i = 0;i<timePoints.size();i++){
            String[] firstTime = timePoints.get(i).split("\\:");
            int firstM = (Integer.valueOf(firstTime[0]) *60+ Integer.valueOf(firstTime[1]));
            for(int j = i+1;j<timePoints.size();j++){        
                
                String[] secondTime = timePoints.get(j).split("\\:");
                int secondM = (Integer.valueOf(secondTime[0]) *60+ Integer.valueOf(secondTime[1]));
                int val = (secondM-firstM+1440)%1440;
                int min = Math.min( val,1440-val);
                result = Math.min(result,min);
            }
        }
        return result;
    }
}

在貼一個hash的做法
public class Solution {
    public int findMinDifference(List<String> timePoints) {
        boolean[] mark = new boolean[24 * 60];
        for (String time : timePoints) {
            String[] t = time.split(":");
            int h = Integer.parseInt(t[0]);
            int m = Integer.parseInt(t[1]);
            if (mark[h * 60 + m]) return 0;
            mark[h * 60 + m] = true;
        }
        
        int prev = 0, min = Integer.MAX_VALUE;
        int first = Integer.MAX_VALUE, last = Integer.MIN_VALUE;
        for (int i = 0; i < 24 * 60; i++) {
            if (mark[i]) {
                if (first != Integer.MAX_VALUE) {
                    min = Math.min(min, i - prev);
                }
                first = Math.min(first, i);
                last = Math.max(last, i);
                prev = i;
            }
        }
        
        min = Math.min(min, (24 * 60 - last + first));
        
        return min;
    }
}

2017/03/28