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 <= persons.length = times.length <= 5000

0 <= persons[i] <= persons.length

times is a strictly increasing array with all elements in [0, 10^9].

TopVotedCandidate.q is called at most 10000 times per test case.

TopVotedCandidate.q(int t) is always called with t >= times[0].

所以這個題使用List儲存每個時間點對應的當前的獲得票數最多的person。在q(t)中，使用二分查詢到第一個小於t的times位置，然後返回這個位置對應的時間得票最多的person即可。

平均的時間複雜度是O(logn)，空間複雜度是O(N).

class TopVotedCandidate {

    //persons [0,1,1,0,0,1,0]

    //times [0,5,10,15,20,25,30]

    List<int[]> list;

    public TopVotedCandidate(int[] persons, int[] times) {

        list = new ArrayList<>();

        Map<Integer, Integer> map = new HashMap<>();

        int lead = -1;

        int leadP = -1;

        for (int i = 0; i < persons.length; i++){

            map.put(persons[i], map.getOrDefault(persons[i],0)+1);

            int[] pair = new int[2];

            pair[0] = times[i];

            if(map.get(persons[i]) >= lead){

                lead = map.get(persons[i]);

                leadP=persons[i];

            }

            pair[1] = leadP;

            list.add(pair);

        }

    }

    public int q(int t) {

        int l = 0;

        int r = list.size()-1;

        while(l <= r){

            int mid = l + (r-l)/2;

            if(t == list.get(mid)[0]){

                return list.get(mid)[1];

            }

            else if(t < list.get(mid)[0]){

                r = mid-1;

            }

            else{

                l = mid+1;

            }

        }

        return list.get(r)[1];

    }

}

/**

 * Your TopVotedCandidate object will be instantiated and called as such:

 * TopVotedCandidate obj = new TopVotedCandidate(persons, times);

 * int param_1 = obj.q(t);

 */