zoukankan      html  css  js  c++  java
  • LeetCode 1172. Dinner Plate Stacks

    原题链接在这里:https://leetcode.com/problems/dinner-plate-stacks/

    题目:

    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 pushpop, and popAtStack.

    题解:

    Have a minHeap to track previous pop at index position.

    When pop at index, check if index is within list size and pointing stack is not empty, add this index to min heap.

    When push, first check if min heap peek is already >= list size. If yes, that means the smallest index is already larger than list size. Those stack must be empty by pop. Clear the min heap.

    Otherwise, check if min heap is empty, if not, add to the last stack, otherwise, add to that index.

    When pop, already clear the empty stack from last and then pop the last non-empty stack.

    Time Complexity: push, O(1). pop, amortize O(list.size()). popAtIndex O(logn). n = minHeap.size().

    Space: O(m). m = list.size() * capacity.

    AC Java: 

     1 class DinnerPlates {
     2     List<Stack<Integer>> list;
     3     PriorityQueue<Integer> minHeap;
     4     int capa;
     5     
     6     public DinnerPlates(int capacity) {
     7         list = new ArrayList<>();
     8         minHeap = new PriorityQueue<>();
     9         capa = capacity;
    10     }
    11     
    12     public void push(int val) {
    13         if(!minHeap.isEmpty() && minHeap.peek() >= list.size()){
    14             minHeap.clear();
    15         }
    16         
    17         if(minHeap.isEmpty()){
    18             if(list.size() == 0 || list.get(list.size() - 1).size() == capa){
    19                 list.add(new Stack<>());
    20             }
    21             
    22             list.get(list.size() - 1).push(val);
    23         }else{
    24             list.get(minHeap.poll()).add(val);
    25         }
    26     }
    27     
    28     public int pop() {
    29         while(list.size() > 0 && list.get(list.size() - 1).size() == 0){
    30             list.remove(list.size() - 1);
    31         }
    32         
    33         if(list.size() == 0){
    34             return -1;
    35         }
    36         
    37         return list.get(list.size() - 1).pop();
    38     }
    39     
    40     public int popAtStack(int index) {
    41         if(index >= list.size()){
    42             return -1;
    43         }
    44         
    45         if(list.get(index).size() == 0){
    46             return -1;
    47         }
    48         
    49         minHeap.add(index);
    50         return list.get(index).pop();
    51     }
    52 }
    53 
    54 /**
    55  * Your DinnerPlates object will be instantiated and called as such:
    56  * DinnerPlates obj = new DinnerPlates(capacity);
    57  * obj.push(val);
    58  * int param_2 = obj.pop();
    59  * int param_3 = obj.popAtStack(index);
    60  */
  • 相关阅读:
    剑指offer--03.从尾到头打印链表
    剑指offer--02.替换空格
    剑指offer--01.二维数组中的查找
    JAVA日记之mybatis-3一对一,一对多,多对多xml与注解配置
    SpringBoot 2.x 自定义拦截器并解决静态资源访问被拦截问题
    springboot项目WEB-INF 目录 jsp页面报404
    Spring Boot 配置拦截器方式
    通过idea创建Maven项目整合Spring+spring mvc+mybatis
    idea创建maven项目
    PLSQL操作Oracle创建用户和表
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/12191642.html
Copyright © 2011-2022 走看看