LeetCode 1381. 设计一个支持增量操作的栈
Table of Contents
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。
