911. 在线选举

合集 - LeetCode 二分法(17)

1.35. 搜索插入位置2024-09-092.704. 二分查找2024-09-093.744. 寻找比目标字母大的最小字母2024-09-094.2529. 正整数和负整数的最大计数2024-09-095.1385. 两个数组间的距离值2024-09-096.2300. 咒语和药水的成功对数2024-09-097.2389. 和有限的最长子序列2024-09-098.1170. 比较字符串最小字母出现频次2024-09-099.2080. 区间内查询数字的频率2024-09-0910.2563. 统计公平数对的数目2024-09-0911.981. 基于时间的键值存储2024-09-1012.1146. 快照数组2024-09-1013.658. 找到 K 个最接近的元素2024-09-1014.1818. 绝对差值和2024-09-10

15.911. 在线选举2024-09-10

16.LCP 08. 剧情触发时间2024-09-1017.[模板题] - 300. 最长递增子序列2024-09-13

收起

题目链接	911. 在线选举
思路	二分
题解链接	[Python/Java/JavaScript/Go] 二分查找
关键点	理解题意：预处理 - 按照times得出每个离散时刻下获胜者的person；询问 - 二分查找到>t的前一个时刻获胜者
时间复杂度	$O(n)$
空间复杂度	$O(n)$

代码实现：

class TopVotedCandidate:
 
    def __init__(self, persons: List[int], times: List[int]):
        n = len(times)
        self.times = times
        self.answer = [-1] * n
 
        cnts = defaultdict(int)
        winner = None
        for i, person in enumerate(persons):
            cnts[person] += 1
            if winner is None or cnts[person] >= cnts[winner]:
                winner = person
            self.answer[i] = winner
        self.n = n
            
    def q(self, t: int) -> int:
        left, right = -1, self.n
        while left + 1 < right:
            mid = (left+right) // 2
            if self.times[mid] <= t:
                left = mid
            else:
                right = mid
        return self.answer[right - 1]

Python-标准库

class TopVotedCandidate:
 
    def __init__(self, persons: List[int], times: List[int]):
        n = len(times)
        self.times = times
        self.answer = [-1] * n
 
        cnts = defaultdict(int)
        winner = None
        for i, person in enumerate(persons):
            cnts[person] += 1
            if winner is None or cnts[person] >= cnts[winner]:
                winner = person
            self.answer[i] = winner
            
    def q(self, t: int) -> int:
        return self.answer[
            bisect_right(self.times, t) - 1
        ]