895. Maximum Frequency Stack

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Implement FreqStack, a class which simulates the operation of a stack-like data structure.

FreqStack has two functions:

• push(int x), which pushes an integer x onto the stack.
• pop(), which removes and returns the most frequent element in the stack.
  • If there is a tie for most frequent element, the element closest to the top of the stack 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]
Explanation:
After making six .push operations, the stack is [5,7,5,7,4,5] from bottom to top.  Then:

pop() -> returns 5, as 5 is the most frequent.
The stack becomes [5,7,5,7,4].

pop() -> returns 7, as 5 and 7 is the most frequent, but 7 is closest to the top.
The stack becomes [5,7,5,4].

pop() -> returns 5.
The stack becomes [5,7,4].

pop() -> returns 4.
The stack becomes [5,7].

Note:

• Calls to FreqStack.push(int x) will be such that 0 <= x <= 10^9.
• It is guaranteed that FreqStack.pop() won't be called if the stack has zero elements.
• The total number of FreqStack.push calls will not exceed 10000 in a single test case.
• The total number of FreqStack.pop calls will not exceed 10000 in a single test case.
• The total number of FreqStack.push and FreqStack.pop calls will not exceed 150000 across all test cases.

一开始想到的就是优先队列+Map，优先队列放（-count, -index, key），但是有点卡住，就是如果key现在count是val，再来一个key，count变成val+1，那岂不是要把之前对应的tuple从优先队列里面删除？但是我不知道index啊，那就还得额外用一个Map来缓存一下index？？想来想去还是太复杂了

然后发现，其实完全不用删除啊！因为你删除的话，后面pop又得加回来啊，还不如就全部放在优先队列里面呢！

class FreqStack:

    def __init__(self):
        self.heap = []
        self.m = collections.defaultdict(int)
        self.counter = 0
        
    def push(self, x):
        self.m[x]+=1
        heapq.heappush(self.heap,(-self.m[x], -self.counter, x))
        self.counter+=1
    
    def pop(self):
        toBeRemoved = heapq.heappop(self.heap)
        self.m[toBeRemoved[2]]-=1
        return toBeRemoved[2]

另外一个解法也是：key的count为val，那就把1..val累加的过中所有的都加到Map里面

    def __init__(self):
        self.freq = collections.Counter()
        self.m = collections.defaultdict(list)
        self.maxf = 0

    def push(self, x):
        freq, m = self.freq, self.m
        freq[x] += 1
        self.maxf = max(self.maxf, freq[x])
        m[freq[x]].append(x)

    def pop(self):
        freq, m, maxf = self.freq, self.m, self.maxf
        x = m[maxf].pop()
        if not m[maxf]: self.maxf = maxf - 1
        freq[x] -= 1
        return x