我的解法：
1.考慮極端情況，無(wú)人就座，只有1人就座且（坐在左邊，中間，最右邊），只剩下最左邊和最右邊的座位等等
2.想的是查找最大的間隔，如果間隔兩端都有人那么就坐中間，如果間隔左邊沒(méi)人坐左邊，左邊有人右邊無(wú)人坐右邊
3.如果間隔大于1/2總長(zhǎng)度就可以跳出了，后面的間隔不會(huì)更大了

class ExamRoom:def __init__(self, n: int):self.seat_list=[]self.seat_set=set()self.left=0self.right=0self.n=ndef seat(self) -> int:n=self.nif self.seat_list==[]:self.seat_list.append(0)self.seat_set.add(0)return 0if len(self.seat_set)>=n-2 and 0 not in self.seat_set:self.seat_list.insert(0,0)self.seat_set.add(0)return 0 last=-1max_gap=n-1-self.seat_list[-1]-1if max_gap<0:max_gap=0left=-1right=nhalf_gap=n//2 insert_loc=0for student in range(0,len(self.seat_list)):gap=self.seat_list[student]-lastif gap//2>max_gap//2: max_gap=gapleft=lastright=self.seat_list[student]insert_loc=studentif max_gap>half_gap:break last=self.seat_list[student]if left in self.seat_set and right in self.seat_set:new_seat=(left+right)//2self.seat_list.insert(insert_loc,new_seat)self.seat_set.add(new_seat)return new_seatif n-1 not in self.seat_set:self.seat_list.append(n-1)self.seat_set.add(n-1)return n-1 self.seat_list.insert(0,0)self.seat_set.add(0)return 0 def leave(self, p: int) -> None:self.seat_list.remove(p)self.seat_set.remove(p)

標(biāo)準(zhǔn)解法：

class ExamRoom:def __init__(self, n: int): self.heap_info = [] self.last = n - 1heappush(self.heap_info, [-self.last, -1, self.last + 1]) def seat(self) -> int:gap, left, right = heappop(self.heap_info)if left < 0:if right > self.last:heappush(self.heap_info, [-self.last, 0, right])elif right > 1:heappush(self.heap_info, [-(right >> 1), 0, right]) return 0elif right > self.last:if gap < -1:heappush(self.heap_info, [-(-gap >> 1), left, self.last]) return self.lastelse:seat_idx = left - gapif gap < -1:heappush(self.heap_info, [-(-gap >> 1), left, seat_idx]) heappush(self.heap_info, [-((right - seat_idx) >> 1), seat_idx, right])elif left + 3 == right:heappush(self.heap_info, [-1, seat_idx, right]) return seat_idxdef leave(self, p: int) -> None:if p == 0:right = 1for i in range(len(self.heap_info)):if self.heap_info[i][1] == 0:_, _, right = self.heap_info.pop(i)breakheappush(self.heap_info, [-right, -1, right]) elif p == self.last:left = p - 1for i in range(len(self.heap_info)):if self.heap_info[i][2] == self.last:_, left, _ = self.heap_info.pop(i)breakheappush(self.heap_info, [left - self.last, left, self.last + 1]) else:cnt = 0left, right = p - 1, p + 1for i in range(len(self.heap_info) - 1, -1, -1):if self.heap_info[i][1] == p:_, _, right = self.heap_info.pop(i)if cnt == 1:breakcnt = 1 elif self.heap_info[i][2] == p:_, left, _ = self.heap_info.pop(i)if cnt == 1:breakcnt = 1 if left < 0:gap = rightelif right > self.last:gap = self.last - leftelse:gap = (right - left) >> 1heappush(self.heap_info, [-gap, left, right]) return None

python的heappush和heappop:
1.heappush(heap,item)建立大、小根堆
heapq.heappush()是往堆中添加新值，此時(shí)自動(dòng)建立了小根堆
不能直接建立大跟堆，所以每次push時(shí)給元素加一個(gè)負(fù)號(hào)（即取相反數(shù)），此時(shí)最小值變最大值，反之亦然，那么實(shí)際上的最大值就可以處于堆頂了，返回時(shí)再取負(fù)即可。
2.heapq.heappop()從堆中彈出并返回最小的值
普通list（即并沒(méi)有進(jìn)行heapify等操作的list），對(duì)他進(jìn)行heappop操作并不會(huì)彈出list中最小的值，而是彈出第一個(gè)值。
對(duì)于小跟堆，會(huì)依次彈出最小的值。

x>>1相當(dāng)于快速的x//2

原文鏈接：https://blog.csdn.net/qq_38022469/article/details/123851001

其他解答：
區(qū)間的左右兩邊用有序列表存儲(chǔ)，元素是一個(gè)元組，排序的key是計(jì)算距離的負(fù)數(shù)，這樣可以確保大的值排在前面，第二個(gè)排序依據(jù)是x[0]就是區(qū)間的左邊位置，確保先坐序號(hào)小的
這樣seat的時(shí)候就相當(dāng)于刪除了原來(lái)的區(qū)間，把其分割為兩個(gè)新的區(qū)間。
兩個(gè)字典存放的是當(dāng)前學(xué)生左右兩邊的學(xué)生，方便刪除的時(shí)候合并區(qū)間，這個(gè)比較好懂一些

from sortedcontainers import SortedListclass ExamRoom:def __init__(self, n: int):def dist(x):l, r = xreturn r - l - 1 if l == -1 or r == n else (r - l) >> 1self.n = nself.ts = SortedList(key=lambda x: (-dist(x), x[0]))self.left = {}self.right = {}self.add((-1, n))def seat(self) -> int:s = self.ts[0]p = (s[0] + s[1]) >> 1if s[0] == -1:p = 0elif s[1] == self.n:p = self.n - 1self.delete(s)self.add((s[0], p))self.add((p, s[1]))return pdef leave(self, p: int) -> None:l, r = self.left[p], self.right[p]self.delete((l, p))self.delete((p, r))self.add((l, r))def add(self, s):self.ts.add(s)self.left[s[1]] = s[0]self.right[s[0]] = s[1]def delete(self, s):self.ts.remove(s)self.left.pop(s[1])self.right.pop(s[0])

作者：lcbin
鏈接：https://leetcode.cn/problems/exam-room/solution/by-lcbin-tstp/
來(lái)源：力扣（LeetCode）

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