LeetCode-1604. 警告一小时内使用相同员工卡大于等于三次的人

题目: https://leetcode-cn.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/

解法一: 暴力解法

根据题目的意思, 每个人和对应时间的关系类似于Map的关系(一对多), 所以, 该题使用哈希表可以解决.

import java.math.BigDecimal;

class Solution {
    public List alertNames(String[] keyName, String[] keyTime) {
        List ans = new ArrayList<>();
        Map> namesTime = new HashMap<>();
        int len = keyName.length;
        for (int i = 0; i < len; i++) {
            String currTime = keyTime[i];
            if (!namesTime.containsKey(keyName[i])) {
                List tmp = new ArrayList<>();
                tmp.add(timeToFloat(keyTime[i]));
                namesTime.put(keyName[i], tmp);
                continue;
            }
            namesTime.computeIfPresent(keyName[i], (s, floats) -> {
                floats.add(timeToFloat(currTime));
                return floats;
            });
        }
        for (Map.Entry entry : namesTime.entrySet()) {
            List value = (List) entry.getValue();
            value.sort((o1, o2) -> Float.compare(o1, o2));
            for (int i = 0; i < value.size() - 2; i++) {
                // 注意: 如果直接浮点数相减会不准确, 这里需要保留2位小数
                BigDecimal bigDecimal = new BigDecimal(value.get(i) - value.get(i + 2));
                Float period = Math.abs(bigDecimal.setScale(2, BigDecimal.ROUND_HALF_UP).floatValue());
                if (period.compareTo(1f) <= 0) {
                    ans.add((String) entry.getKey());
                    break;
                }
            }
        }
        ans.sort(Comparator.naturalOrder());
        return ans;
    }

    private float timeToFloat(String time) {
        return Float.valueOf(time.replace(":", "."));
    }
}