[LeetCode] 911. Online Election

SoapEye 发布于2019-08-16 13:55 / 675人阅读

Problem

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].

Solution

class TopVotedCandidate {

    Map history = new HashMap<>();
    int[] times;
    public TopVotedCandidate(int[] persons, int[] times) {
        this.times = times;
        Map votes = new HashMap<>();
        int n = persons.length;
        int leader = persons[0];
        for (int i = 0; i < n; i++) {
            votes.put(persons[i], votes.getOrDefault(persons[i], 0)+1);
            if (votes.get(persons[i]) >= votes.get(leader)) leader = persons[i];
            history.put(times[i], leader);
        }
    }
    
    public int q(int t) {
        int i = Arrays.binarySearch(times, t);
        if (i < 0) return history.get(times[-i-2]);
        return history.get(times[i]);
    }
}

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