design-a-stack-with-increment-operation

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Algorithms

Design a stack that supports increment operations on its elements.

Implement the <code>CustomStack</code> class:

<ul>
	<li><code>CustomStack(int maxSize)</code> Initializes the object with <code>maxSize</code> which is the maximum number of elements in the stack.</li>
	<li><code>void push(int x)</code> Adds <code>x</code> to the top of the stack if the stack has not reached the <code>maxSize</code>.</li>
	<li><code>int pop()</code> Pops and returns the top of the stack or <code>-1</code> if the stack is empty.</li>
	<li><code>void inc(int k, int val)</code> Increments the bottom <code>k</code> elements of the stack by <code>val</code>. If there are less than <code>k</code> elements in the stack, increment all the elements in the stack.</li>
</ul>

&nbsp;
Example 1:

<pre>
Input
[&quot;CustomStack&quot;,&quot;push&quot;,&quot;push&quot;,&quot;pop&quot;,&quot;push&quot;,&quot;push&quot;,&quot;push&quot;,&quot;increment&quot;,&quot;increment&quot;,&quot;pop&quot;,&quot;pop&quot;,&quot;pop&quot;,&quot;pop&quot;]
[[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 stk = new CustomStack(3); // Stack is Empty []
stk.push(1); // stack becomes [1]
stk.push(2); // stack becomes [1, 2]
stk.pop(); // return 2 --&gt; Return top of the stack 2, stack becomes [1]
stk.push(2); // stack becomes [1, 2]
stk.push(3); // stack becomes [1, 2, 3]
stk.push(4); // stack still [1, 2, 3], Do not add another elements as size is 4
stk.increment(5, 100); // stack becomes [101, 102, 103]
stk.increment(2, 100); // stack becomes [201, 202, 103]
stk.pop(); // return 103 --&gt; Return top of the stack 103, stack becomes [201, 202]
stk.pop(); // return 202 --&gt; Return top of the stack 202, stack becomes [201]
stk.pop(); // return 201 --&gt; Return top of the stack 201, stack becomes []
stk.pop(); // return -1 --&gt; Stack is empty return -1.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= maxSize, x, k &lt;= 1000</code></li>
	<li><code>0 &lt;= val &lt;= 100</code></li>
	<li>At most <code>1000</code> calls will be made to each method of <code>increment</code>, <code>push</code> and <code>pop</code> each separately.</li>
</ul>

Design a Stack With Increment Operation

Stack

Design

Array

请你设计一个支持对其元素进行增量操作的栈。

实现自定义栈类 <code>CustomStack</code> ：

<ul>
	<li><code>CustomStack(int maxSize)</code>：用 <code>maxSize</code> 初始化对象，<code>maxSize</code> 是栈中最多能容纳的元素数量。</li>
	<li><code>void push(int x)</code>：如果栈还未增长到 <code>maxSize</code> ，就将 <code>x</code> 添加到栈顶。</li>
	<li><code>int pop()</code>：弹出栈顶元素，并返回栈顶的值，或栈为空时返回 -1 。</li>
	<li><code>void inc(int k, int val)</code>：栈底的 <code>k</code> 个元素的值都增加 <code>val</code> 。如果栈中元素总数小于 <code>k</code> ，则栈中的所有元素都增加 <code>val</code> 。</li>
</ul>

&nbsp;

示例：

<pre>
输入：
["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]
输出：
[null,null,null,2,null,null,null,null,null,103,202,201,-1]
解释：
CustomStack stk = new CustomStack(3); // 栈是空的 []
stk.push(1); // 栈变为 [1]
stk.push(2); // 栈变为 [1, 2]
stk.pop(); // 返回 2 --&gt; 返回栈顶值 2，栈变为 [1]
stk.push(2); // 栈变为 [1, 2]
stk.push(3); // 栈变为 [1, 2, 3]
stk.push(4); // 栈仍然是 [1, 2, 3]，不能添加其他元素使栈大小变为 4
stk.increment(5, 100); // 栈变为 [101, 102, 103]
stk.increment(2, 100); // 栈变为 [201, 202, 103]
stk.pop(); // 返回 103 --&gt; 返回栈顶值 103，栈变为 [201, 202]
stk.pop(); // 返回 202 --&gt; 返回栈顶值 202，栈变为 [201]
stk.pop(); // 返回 201 --&gt; 返回栈顶值 201，栈变为 []
stk.pop(); // 返回 -1 --&gt; 栈为空，返回 -1
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= maxSize, x, k &lt;= 1000</code></li>
	<li><code>0 &lt;= val &lt;= 100</code></li>
	<li>每种方法 <code>increment</code>，<code>push</code> 以及 <code>pop</code> 分别最多调用 <code>1000</code> 次</li>
</ul>