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]
.
题目理解:
给定一系列时间节点和投票信息,问给定时刻谁的票数最多
解题思路:
按事件顺序,找到每一个时间节点处票数最多的人,用TreeMap存储时间和候选人代号,查询的时候查找最接近但是小于当前时刻的时间节点,返回候选人信息
代码 如下:
class TopVotedCandidate {
TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
public TopVotedCandidate(int[] persons, int[] times) {
int max = 0, len = persons.length;
int[] count = new int[len + 1];
for(int i = 0; i < len; i++) {
int person = persons[i], time = times[i];
count[person]++;
if(count[max] <= count[person])
max = person;
map.put(time, max);
}
}
public int q(int t) {
return map.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);
*/