LeetCode 1381. 设计一个支持增量操作的栈

Table of Contents

一、中文版

二、英文版

三、My answer

四、解题报告

 

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一、中文版

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

 

二、英文版

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
At most 1000 calls will be made to each method of increment, push and pop each separately.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/design-a-stack-with-increment-operation
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

三、My answer

class CustomStack:

    def __init__(self, maxSize: int):
        self.size = maxSize
        self.array = []

    def push(self, x: int) -> None:
        if len(self.array) < self.size:
            self.array.append(x)

    def pop(self) -> int:
        if len(self.array) != 0:
            r = self.array.pop()
            return r
        else:
            return -1


    def increment(self, k: int, val: int) -> None:
        for i in range(min(k, len(self.array))):
            self.array[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)

四、解题报告

1、用 list 实现栈

2、一开始想错误，想用类似 array = [0] * maxSize 的形式，可是赋初值后又不对。其实只要用变量来接收传进来的参数 maxSize，再用 size 作为限制就行。