LeetCode 1172. Dinner Plate Stacks 設計中間彈棧

You have an infinite number of stacks arranged in a row and numbered (left to right) from 0, each of the stacks has the same maximum capacity.

Implement the DinnerPlates class:

• DinnerPlates(int capacity) Initializes the object with the maximum capacity of the stacks.
• void push(int val) pushes the given positive integer val into the leftmost stack with size less than capacity.
• int pop() returns the value at the top of the rightmost non-empty stack and removes it from that stack, and returns -1 if all stacks are empty.
• int popAtStack(int index) returns the value at the top of the stack with the given index and removes it from that stack, and returns -1 if the stack with that given index is empty.

Example:

Input: 
["DinnerPlates","push","push","push","push","push","popAtStack","push","push","popAtStack","popAtStack","pop","pop","pop","pop","pop"]
[[2],[1],[2],[3],[4],[5],[0],[20],[21],[0],[2],[],[],[],[],[]]
Output: 
[null,null,null,null,null,null,2,null,null,20,21,5,4,3,1,-1]

Explanation: 
DinnerPlates D = DinnerPlates(2);  // Initialize with capacity = 2
D.push(1);
D.push(2);
D.push(3);
D.push(4);
D.push(5);         // The stacks are now:  2  4
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.popAtStack(0);   // Returns 2.  The stacks are now:     4
                                                       1  3  5
                                                       ﹈ ﹈ ﹈
D.push(20);        // The stacks are now: 20  4
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.push(21);        // The stacks are now: 20  4 21
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.popAtStack(0);   // Returns 20.  The stacks are now:     4 21
                                                        1  3  5
                                                        ﹈ ﹈ ﹈
D.popAtStack(2);   // Returns 21.  The stacks are now:     4
                                                        1  3  5
                                                        ﹈ ﹈ ﹈ 
D.pop()            // Returns 5.  The stacks are now:      4
                                                        1  3 
                                                        ﹈ ﹈  
D.pop()            // Returns 4.  The stacks are now:   1  3 
                                                        ﹈ ﹈   
D.pop()            // Returns 3.  The stacks are now:   1 
                                                        ﹈   
D.pop()            // Returns 1.  There are no stacks.
D.pop()            // Returns -1.  There are still no stacks.

 

Constraints:

• 1 <= capacity <= 20000
• 1 <= val <= 20000
• 0 <= index <= 100000
• At most 200000 calls will be made to push, pop, and popAtStack.

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這題用堆並不難想到，難點是如何簡化codes，有下面幾個問題需要考慮清楚：

1. 堆裏放什麼？所有刪過的棧下標都放進堆裏。可能存在stack[stack]

2. 利用push_at和pop_at來簡化代碼

3. size,idx這些成員變量沒有必要記，python的len複雜度是O(1)

import heapq
from collections import defaultdict
class DinnerPlates:

    def __init__(self, capacity: int):
        self.cap = capacity
        self.pq,self.sta=[],[]
        
    def push(self, val: int) -> None:
        def push_at(index, val):
            if (index < 0 or index >= len(self.sta) or len(self.sta[index]) == self.cap):
                self.sta.append([val])
            else:
                self.sta[index].append(val)
        
        insert_idx = heapq.heappop(self.pq) if self.pq else len(self.sta)-1
        push_at(insert_idx, val)

    def pop(self) -> int:
        while (self.sta and (not self.sta[-1])):
            self.sta.pop()
        return self.popAtStack(len(self.sta)-1)
    
    def popAtStack(self, index: int) -> int:
        if (0 <= index < len(self.sta) and self.sta[index]):
            heapq.heappush(self.pq,index)
            return self.sta[index].pop()
        return -1