文章作者：Tyan
博客：noahsnail.com | CSDN | 简书
1. Description
2. Solution
**解析：**Version 1，创建一个统计年份人数数组，遍历所有日志，遍历出生、死亡之间的年份，累加对应年份的人口，最后找出人口最多最早的年份，注意边界值。Version 2是遍历所有年份，再遍历所有日志，统计每个年份的人口并比较，比Version 1要慢一些。Version 3根据出生年份和死亡年份来更新当年的人口变化，出省年份人口数量加1，死亡年份人口数量减1，最后遍历所有年份，累加每个年份的人口变化即为当前年份的总人口，注意，此时2050年死亡人口要减1，因此边界值要变为end - start + 1。
- Version 1
class Solution:def maximumPopulation(self, logs: List[List[int]]) -> int:start = 1950end = 2050stat = [0] * (end - start)for birth, death in logs:for i in range(birth, death):stat[i - start] += 1temp = 0for index, count in enumerate(stat):if count > temp:result = indextemp = countreturn result + start
- Version 2
class Solution:def maximumPopulation(self, logs: List[List[int]]) -> int:start = 1950end = 2050temp = 0for i in range(start, end):count = 0for birth, death in logs: if birth <= i and i < death:count += 1if count > temp:temp = countresult = ireturn result
- Version 3
class Solution:def maximumPopulation(self, logs: List[List[int]]) -> int:start = 1950end = 2050stat = [0] * (end - start + 1)for birth, death in logs:stat[birth - start] += 1stat[death - start] -= 1temp = 0current = 0for index, count in enumerate(stat):current += countif current > temp:result = indextemp = currentreturn result + start
Reference
- https://leetcode.com/problems/maximum-population-year/