示例 1：
输入：inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]输出：[null,null,"bar","bar",null,"bar2","bar2"]解释：  TimeMap kv;   kv.set("foo", "bar", 1); // 存储键 "foo" 和值 "bar" 以及时间戳 timestamp = 1   kv.get("foo", 1);  // 输出 "bar"   kv.get("foo", 3); // 输出 "bar" 因为在时间戳 3 和时间戳 2 处没有对应 "foo" 的值，所以唯一的值位于时间戳 1 处（即 "bar"）   kv.set("foo", "bar2", 4);   kv.get("foo", 4); // 输出 "bar2"   kv.get("foo", 5); // 输出 "bar2"  
来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/time-based-key-value-store著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class TimeMap {
    /** Initialize your data structure here. */    Map<String, Map<Integer, String>> map;    public TimeMap() {         map = new HashMap<>();    }        public void set(String key, String value, int timestamp) {        Map<Integer, String> temp;        if (map.containsKey(key)) {            temp = map.get(key);        } else {            temp = new HashMap<>();        }        temp.put(timestamp, value);
        map.put(key, temp);    }        public String get(String key, int timestamp) {        String ans = "";        if (map.containsKey(key)) {            Map<Integer, String> temp = map.get(key);            while (timestamp >= 0) {                if (temp.containsKey(timestamp)) {                    ans = temp.get(timestamp);                    break;                }                timestamp--;            }        }         return ans;    }}


/** * Your TimeMap object will be instantiated and called as such: * TimeMap obj = new TimeMap(); * obj.set(key,value,timestamp); * String param_2 = obj.get(key,timestamp); */