Design a stack which supports the following operations.

Implement theCustomStackclass:

• CustomStack(int maxSize)Initializes the object withmaxSizewhich is the maximum number of elements in the stack or do nothing if the stack reached themaxSize.
• void push(int x)Addsxto the top of the stack if the stack hasn't reached themaxSize
  .
• int pop()Pops and returns the top of stack or-1if the stack is empty.
• void inc(int k, int val)Increments the bottomkelements of the stack byval. If there are less thankelements in the stack, just increment all the elements in the stack.

Example 1:

Input
["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]
Output
[null,null,null,2,null,null,null,null,null,103,202,201,-1]
Explanation
CustomStack customStack = new CustomStack(3); // Stack is Empty []
customStack.push(1);                          // stack becomes [1]
customStack.push(2);                          // stack becomes [1, 2]
customStack.pop();                            // return 2 --> Return top of the stack 2, stack becomes [1]
customStack.push(2);                          // stack becomes [1, 2]
customStack.push(3);                          // stack becomes [1, 2, 3]
customStack.push(4);                          // stack still [1, 2, 3], Don't add another elements as size is 4
customStack.increment(5, 100);                // stack becomes [101, 102, 103]
customStack.increment(2, 100);                // stack becomes [201, 202, 103]
customStack.pop();                            // return 103 --> Return top of the stack 103, stack becomes [201, 202]
customStack.pop();                            // return 202 --> Return top of the stack 102, stack becomes [201]
customStack.pop();                            // return 201 --> Return top of the stack 101, stack becomes []
customStack.pop();                            // return -1 --> Stack is empty return -1.

Constraints:

• 1 <= maxSize <= 1000
• 1 <= x <= 1000
• 1 <= k <= 1000
• 0 <= val <= 100
• At most1000calls will be made to each method ofincrement,pushandpopeach separately.

設計一個支援增量操作的棧。

請你設計一個支援下述操作的棧。

實現自定義棧類 CustomStack ：

CustomStack(int maxSize)：用 maxSize 初始化物件，maxSize 是棧中最多能容納的元素數量，棧在增長到 maxSize 之後則不支援 push 操作。
void push(int x)：如果棧還未增長到 maxSize ，就將 x 新增到棧頂。
int pop()：彈出棧頂元素，並返回棧頂的值，或棧為空時返回 -1 。
void inc(int k, int val)：棧底的 k 個元素的值都增加 val 。如果棧中元素總數小於 k ，則棧中的所有元素都增加 val 。

來源：力扣（LeetCode）
連結：https://leetcode-cn.com/problems/design-a-stack-with-increment-operation
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這是一道設計題，除了支援正常的關於棧的操作，還多一個功能，支援增量操作。這道題我給出兩種做法，一種是用list模擬stack，另一種是利用一個額外陣列去記錄增量。

首先是用list模擬stack，空間O(n)。

push O(1) - 只要stack的size小於maxSize，就可以往stack增加元素

pop O(1) - 只要stack裡面還有東西，就可以往外pop，否則就返回-1

increment O(n) - 用一個for迴圈去強行修改list裡面對應下標範圍內的元素

 1 class CustomStack {
 2     private List<Integer> stack = new ArrayList<>();
 3     private int size;
 4 
 5     public CustomStack(int maxSize) {
 6         size = maxSize;
 7     }
 8 
 9     public void push(int x) {
10         if (stack.size() < size) {
11             stack.add(x);
12         }
13     }
14 
15     public int pop() {
16         return stack.isEmpty() ? -1 : stack.remove(stack.size() - 1);
17     }
18 
19     public void increment(int k, int val) {
20         for (int i = 0; i < k && i < stack.size(); i++) {
21             stack.set(i, stack.get(i) + val);
22         }
23     }
24 }
25 
26 /**
27  * Your CustomStack object will be instantiated and called as such:
28  * CustomStack obj = new CustomStack(maxSize);
29  * obj.push(x);
30  * int param_2 = obj.pop();
31  * obj.increment(k,val);
32  */

再來是利用一個額外陣列去記錄增量。這裡我著重解釋一下increment函式，首先我們需要一個和maxSize等長的陣列來記錄stack裡面每個元素可能的增量，介於K有可能大於stack目前的size，所以在做increment的時候需要在K和stack.size()之間取較小值。當需要改動的數字個數 i > 0的時候，我們需要把增量val加到陣列對應的下標 i - 1 上。這樣當我們做pop操作的時候，可以順帶取到這個增量add[i]。同時因為增量影響的是下標從0到 i - 1範圍內的所有數字，所以當i > 0的時候，add[i] 上的增量val也需要累加到 add[i - 1] 上。

空間O(n)

push O(1) - 只要stack的size小於maxSize，就可以往stack增加元素

pop O(1) - 只要stack裡面還有東西，就可以往外pop，否則就返回-1

increment O(1) - 用一個for迴圈去強行修改list裡面對應下標範圍內的元素

 1 class CustomStack {
 2     private int n;
 3     int[] add;
 4     Stack<Integer> stack;
 5 
 6     public CustomStack(int maxSize) {
 7         n = maxSize;
 8         add = new int[n];
 9         stack = new Stack<>();
10     }
11 
12     public void push(int x) {
13         if (stack.size() < n) {
14             stack.push(x);
15         }
16     }
17 
18     public int pop() {
19         int i = stack.size() - 1;
20         if (i < 0) {
21             return -1;
22         }
23         if (i > 0) {
24             add[i - 1] += add[i];
25         }
26         int res = stack.pop() + add[i];
27         add[i] = 0;
28         return res;
29     }
30 
31     public void increment(int k, int val) {
32         int i = Math.min(k, stack.size());
33         if (i > 0) {
34             add[i - 1] += val;
35         }
36     }
37 }
38 
39 /**
40  * Your CustomStack object will be instantiated and called as such:
41  * CustomStack obj = new CustomStack(maxSize);
42  * obj.push(x);
43  * int param_2 = obj.pop();
44  * obj.increment(k,val);
45  */

LeetCode 題目總結