请你设计一个支持下述操作的栈。
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,栈在增长到maxSize
之后则不支持push
操作。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
。
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 customStack = new CustomStack(3); // 栈是空的 [] customStack.push(1); // 栈变为 [1] customStack.push(2); // 栈变为 [1, 2] customStack.pop(); // 返回 2 --> 返回栈顶值 2,栈变为 [1] customStack.push(2); // 栈变为 [1, 2] customStack.push(3); // 栈变为 [1, 2, 3] customStack.push(4); // 栈仍然是 [1, 2, 3],不能添加其他元素使栈大小变为 4 customStack.increment(5, 100); // 栈变为 [101, 102, 103] customStack.increment(2, 100); // 栈变为 [201, 202, 103] customStack.pop(); // 返回 103 --> 返回栈顶值 103,栈变为 [201, 202] customStack.pop(); // 返回 202 --> 返回栈顶值 202,栈变为 [201] customStack.pop(); // 返回 201 --> 返回栈顶值 201,栈变为 [] customStack.pop(); // 返回 -1 --> 栈为空,返回 -1
1 <= maxSize <= 1000
1 <= x <= 1000
1 <= k <= 1000
0 <= val <= 100
- 每种方法
increment
,push
以及pop
分别最多调用1000
次
struct CustomStack {
stack: Vec<i32>,
max_size: usize,
}
/**
* `&self` means the method takes an immutable reference.
* If you need a mutable reference, change it to `&mut self` instead.
*/
impl CustomStack {
fn new(maxSize: i32) -> Self {
Self {
stack: Vec::with_capacity(maxSize as usize),
max_size: maxSize as usize,
}
}
fn push(&mut self, x: i32) {
if self.stack.len() < self.max_size {
self.stack.push(x);
}
}
fn pop(&mut self) -> i32 {
self.stack.pop().unwrap_or(-1)
}
fn increment(&mut self, k: i32, val: i32) {
for i in 0..(k as usize).min(self.stack.len()) {
self.stack[i] += val;
}
}
}
/**
* Your CustomStack object will be instantiated and called as such:
* let obj = CustomStack::new(maxSize);
* obj.push(x);
* let ret_2: i32 = obj.pop();
* obj.increment(k, val);
*/