LeetCode 1381. Design a Stack With Increment Operation设计一个支持增量操作的栈【Medium】【Python】【栈】
LeetCode
Design a stack which supports the following operations.
Implement the CustomStack
class:
CustomStack(int maxSize)
Initializes the object with maxSize
which is the maximum number of elements in the stack or do nothing if the stack reached the maxSize
.void push(int x)
Adds x
to the top of the stack if the stack hasn’t reached the maxSize
.int pop()
Pops and returns the top of 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, 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
1000
calls will be made to each method of increment
, push
and pop
each separately.力扣
请你设计一个支持下述操作的栈。
实现自定义栈类 CustomStack :
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
increment
,push
以及 pop
分别最多调用 1000
次栈
这里我犯了个错,pop 时记得要把栈顶元素弹出,不能只是返回,一定要弹出。
class CustomStack:
def __init__(self, maxSize: int):
self.size = 0
self.maxSize = maxSize
self.customStack = []
def push(self, x: int) -> None:
if self.size < self.maxSize:
self.customStack.append(x)
self.size += 1
def pop(self) -> int:
if self.size == 0:
return -1
temp = self.customStack[self.size - 1]
# 要把栈顶元素弹出去,即删除栈顶元素
del self.customStack[self.size - 1]
self.size -= 1
return temp
def increment(self, k: int, val: int) -> None:
for i in range(min(k, self.size)):
self.customStack[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)
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