1381. 设计一个支持增量操作的栈

🟠   🔖  栈 设计 数组  🔗 力扣 LeetCode

题目

Design a stack that supports increment operations on its elements.

Implement the CustomStack class:

• CustomStack(int maxSize) Initializes the object with maxSize which is the maximum number of elements in the stack.
• void push(int x) Adds x to the top of the stack if the stack has not reached the maxSize.
• int pop() Pops and returns the top of the stack or -1 if the stack is empty.
• void inc(int k, int val) Increments the bottom k elements of the stack by val. If there are less than k elements in the stack, 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 stk = new CustomStack(3); // Stack is Empty []

stk.push(1); // stack becomes [1]

stk.push(2); // stack becomes [1, 2]

stk.pop(); // return 2 --> 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 --> Return top of the stack 103, stack becomes [201, 202]

stk.pop(); // return 202 --> Return top of the stack 202, stack becomes [201]

stk.pop(); // return 201 --> Return top of the stack 201, stack becomes []

stk.pop(); // return -1 --> Stack is empty return -1.

Constraints:

• 1 <= maxSize, x, k <= 1000
• 0 <= val <= 100
• At most 1000 calls will be made to each method of increment, push and pop each separately.

题目大意

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 CustomStack ：

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

解题思路

按照题目要求实现即可：

• 对于 push 操作，首先判断当前元素的个数是否达到上限，如果没有达到，就把 top 后移一个位置并添加一个元素。
• 对于 pop 操作，首先判断当前栈是否为空，非空返回栈顶元素并将 top 前移一位，否则返回 −1。
• 对于 inc 操作，直接对栈底的最多 k 个元素加上 val。

复杂度分析

• 时间复杂度：初始化（构造函数）、push 操作和 pop 操作的渐进时间复杂度为 O(1)，inc 操作的渐进时间复杂度为 O(k)。
• 空间复杂度：这里用到了一个长度为 maxSize 的数组作为辅助空间，渐进空间复杂度为 O(maxSize)。

代码

/**
 * @param {number} maxSize
 */
var CustomStack = function (maxSize) {
	this.stack = [];
	this.max = maxSize;
};

/**
 * @param {number} x
 * @return {void}
 */
CustomStack.prototype.push = function (x) {
	if (this.stack.length < this.max) {
		this.stack.push(x);
	}
};

/**
 * @return {number}
 */
CustomStack.prototype.pop = function () {
	if (this.stack.length > 0) {
		return this.stack.pop();
	}
	return -1;
};

/**
 * @param {number} k
 * @param {number} val
 * @return {void}
 */
CustomStack.prototype.increment = function (k, val) {
	k = Math.min(k, this.stack.length);
	for (let i = 0; i < k; i++) {
		this.stack[i] += val;
	}
};

/**
 * Your CustomStack object will be instantiated and called as such:
 * var obj = new CustomStack(maxSize)
 * obj.push(x)
 * var param_2 = obj.pop()
 * obj.increment(k,val)
 */