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 most10000
times per test case.TopVotedCandidate.q(int t)
is always called witht >= times[0]
.
预先生成好答案。调用q的时候用binary search。这里简单一点用treemap来作个弊
constructor O(n), q O(lg n)
class TopVotedCandidate { private TreeMap<Integer, Integer> winners; public TopVotedCandidate(int[] persons, int[] times) { winners = new TreeMap<>(); Map<Integer, Integer> personToVotes = new HashMap<>(); int winner = -1; personToVotes.put(winner, 0); for (int i = 0; i < persons.length; i++) { int p = persons[i]; personToVotes.put(p, personToVotes.getOrDefault(p, 0) + 1); if (personToVotes.get(p) >= personToVotes.get(winner)) { winner = p; } winners.put(times[i], winner); } } public int q(int t) { return winners.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); */
No comments:
Post a Comment