题目:原题链接(困难)
标签:栈、设计、哈希表
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
---|---|---|---|
Ans 1 (Python) | push = O ( 1 ) O(1) O(1) ; pop = O ( N l o g N ) O(NlogN) O(NlogN) | O ( N ) O(N) O(N) | 超出时间限制 |
Ans 2 (Python) | push = O ( 1 ) O(1) O(1) ; pop = O ( 1 ) O(1) O(1) | O ( N ) O(N) O(N) | 384ms (49.62%) |
Ans 3 (Python) |
LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一(暴力解法):
class FreqStack:
def __init__(self):
self.count = collections.Counter()
self.stack = []
def push(self, x: int) -> None:
self.count[x] += 1
self.stack.append(x)
def pop(self) -> int:
# 计算所有最频繁的元素
most_common = self.count.most_common()
maybe_ans = [most_common[0][0]]
now_max = most_common[0][1]
idx = 1
while idx < len(most_common):
if most_common[idx][1] == now_max:
maybe_ans.append(most_common[idx][0])
idx += 1
else:
break
# 计算最靠近栈顶的元素
for i in range(len(self.stack) - 1, -1, -1):
if self.stack[i] in maybe_ans:
self.count[self.stack[i]] -= 1
return self.stack.pop(i)
解法二(双层栈):
class FreqStack:
def __init__(self):
self.count = collections.Counter()
self.stack = collections.defaultdict(list)
self.max_freq = 0
def push(self, x: int) -> None:
self.count[x] += 1
self.max_freq = max(self.max_freq, self.count[x])
self.stack[self.count[x]].append(x)
def pop(self) -> int:
ans = self.stack[self.max_freq].pop()
if len(self.stack[self.max_freq]) == 0:
self.max_freq -= 1
self.count[ans] -= 1
return ans