In an election, the i-th vote was cast for persons[i] at time times[i].

Now, we would like to implement the following query function: TopVotedCandidate.q(int t) will return the number of the person that was leading the election at time t.  

Votes cast at time t

Example 1:

Input: ["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1,0,0,1,0],[0,5,10,15,20,25,30]],[3],[12],[25],[15],[24],[8]]
Output: [null,0,1,1,0,0,1]
Explanation: 
At time 3, the votes are [0], and 0 is leading.
At time 12, the votes are [0,1,1], and 1 is leading.
At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.)
This continues for 3 more queries at time 15, 24, and 8.

Note:

1. 1 <= persons.length = times.length <= 5000
2. 0 <= persons[i] <= persons.length
3. times is a strictly increasing array with all elements in [0, 10^9].
4. TopVotedCandidate.q is called at most 10000 times per test case.
5. TopVotedCandidate.q(int t) is always called with t >= times[0]
   .

題目理解：

給定一系列時間節點和投票資訊，問給定時刻誰的票數最多

解題思路：

按事件順序，找到每一個時間節點處票數最多的人，用TreeMap儲存時間和候選人代號，查詢的時候查詢最接近但是小於當前時刻的時間節點，返回候選人資訊

程式碼 如下：

class TopVotedCandidate {

	TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
    public TopVotedCandidate(int[] persons, int[] times) {
        int max = 0, len = persons.length;
        int[] count = new int[len + 1];
        for(int i = 0; i < len; i++) {
        	int person = persons[i], time = times[i];
        	count[person]++;
        	if(count[max] <= count[person])
        		max = person;
        	map.put(time, max);
        }
    }
    
    public int q(int t) {
        return map.floorEntry(t).getValue();
    }
}

/**
 * Your TopVotedCandidate object will be instantiated and called as such:
 * TopVotedCandidate obj = new TopVotedCandidate(persons, times);
 * int param_1 = obj.q(t);
 */