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 will count towards our query.  In the case of a tie, the most recent vote (among tied candidates) wins.

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].

題目大意：給定一個person陣列和time陣列，person[i],time[i]意味著在time[i]這個時候有人給person[i]這個人投了一票

要求對於給定的每個時間，找出這個時間得票最多的人。

一開始我寫的很複雜，記錄了每次投票後所有人的總得票，然後對於每個時間都去找小於等於給定時間的最小值。。程式碼寫的很是複雜，結束後我參考了discuss區大佬們的程式碼，豁然開朗。

class TopVotedCandidate {

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

遍歷每次投票，加入到TreeMap中，更新每次投票後目前獲得票數最高的人和其得票，存放於另外一個數組中。這樣還解決了如果票數一樣得找出最近投票那個人這個問題。