# # 【LeetCode】911. Online Election 解题报告（Python）

## # 题目描述：

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 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].

## # 解题方法

``````class TopVotedCandidate(object):

def __init__(self, persons, times):
"""
:type persons: List[int]
:type times: List[int]
"""
self.leads, self.times, count = [], times, {}
for p, t in zip(persons, times):
count[p] = count.get(p, 0) + 1

def q(self, t):
"""
:type t: int
:rtype: int
"""
l, r = 0, len(self.times) - 1
while l <= r:
mid = (l + r) // 2
if self.times[mid] == t:
elif self.times[mid] > t:
r = mid - 1
else:
l = mid + 1

# Your TopVotedCandidate object will be instantiated and called as such:
# obj = TopVotedCandidate(persons, times)
# param_1 = obj.q(t)
``````

https://leetcode.com/problems/online-election/discuss/173382/C++JavaPython-Binary-Search-in-Times

## # 日期

2018 年 9 月 24 日 —— 祝大家中秋节快乐