動機
有趣的一題
Problem
Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack.
Implement the FreqStack
class:
FreqStack()
constructs an empty frequency stack.void push(int val)
pushes an integerval
onto the top of the stack.int pop()
removes and returns the most frequent element in the stack.- If there is a tie for the most frequent element, the element closest to the stack's top is removed and returned.
Example 1:
Input[FreqStack, push, push, push, push, push, push, pop, pop, pop, pop][[], [5], [7], [5], [7], [4], [5], [], [], [], []]Output[null, null, null, null, null, null, null, 5, 7, 5, 4]ExplanationFreqStack freqStack = new FreqStack();freqStack.push(5); // The stack is [5]freqStack.push(7); // The stack is [5,7]freqStack.push(5); // The stack is [5,7,5]freqStack.push(7); // The stack is [5,7,5,7]freqStack.push(4); // The stack is [5,7,5,7,4]freqStack.push(5); // The stack is [5,7,5,7,4,5]freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4].freqStack.pop(); // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4].freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,4].freqStack.pop(); // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7].
Constraints:
0 <= val <= 109
- At most
2 * 104
calls will be made topush
andpop
. - It is guaranteed that there will be at least one element in the stack before calling
pop
.
Sol
class FreqStack:
def __init__(self):
self.stks = defaultdict(list)
self.v2l = defaultdict(int)
self.max_len = 0
def push(self, val: int) -> None:
self.v2l[val] += 1
self.stks[self.v2l[val]].append(val)
self.max_len = max(self.max_len, self.v2l[val])
def pop(self) -> int:
val = self.stks[self.max_len].pop()
if not self.stks[self.max_len]:
self.max_len -= 1
self.v2l[val] -= 1
return val