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()
: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
。
示例:
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
提示:
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
思路:
数据规模在1000以下,所以直接用暴力法就可以了。
高端解似乎是线段树。
时间复杂度:O(N^2)
空间复杂度:O(N)
class CustomStack(object):
def __init__(self, maxSize):
"""
:type maxSize: int
"""
self.stack = []
self.maxSize = maxSize
def push(self, x):
"""
:type x: int
:rtype: None
"""
if len(self.stack) < self.maxSize:
self.stack.append(x)
def pop(self):
"""
:rtype: int
"""
return self.stack.pop() if self.stack else -1
def increment(self, k, val):
"""
:type k: int
:type val: int
:rtype: None
"""
for i in range(min(k, len(self.stack))):
self.stack[i] += val
# Your CustomStack object will be instantiated and called as such:
# obj = CustomStack(maxSize)
# obj.push(x)
# param_2 = obj.pop()
# obj.increment(k,val)