911. Online Election

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].

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Resolve ideas: 

Use Hash Tables to Improve Running Speed

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Code:

class TopVotedCandidate {
    final Map<Integer,Integer> result = new HashMap<>();
    int[] time = null;
    public TopVotedCandidate(int[] persons, int[] times) {
        time = times;
        int[] maxVal = new int[persons.length];
        int max = 0;
        int index = 0;
        for (int i = 0; i < persons.length; i++) {
            int val = maxVal[persons[i]]++;
            if (val>=max){
                max = val;
                index = i;
            }
            result.put(times[i],persons[index]);
        }
    }
    public int q(int t) {
        while (!result.containsKey(t--)){}
        return result.get(++t);
    }
}