leetcode 1381. Design a Stack With Increment Operation（python）

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描述

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.

Note:

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.

解析

根据题意，就是要求自己设计一个栈相关的操作，每个方法都规定了输入、操作方法和输出，思路比较简单：通过列表及其相关的内置函数完成相关的操作即可，只要懂得了栈的概念，实现起来比较简单。

• __init__ 函数中主要是初始化列表 self.stack 和 self.maxSize 两个全局变量
• push 函数主要是在 self.stack 长度不大于 self.maxSize 的情况下将元素压入栈中
• pop 函数主要是当 self.stack 不为空的时候返回栈顶的元素，否则返回 -1
• increment 函数就是将底部的 k 个元素都加上 val ，如果 self.stack 长度小于 k ，则全栈元素都加上 val

解答

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
        """
        if len(self.stack)>0:
            return self.stack.pop()
        return -1

    def increment(self, k, val):
        """
        :type k: int
        :type val: int
        :rtype: None
        """
        if self.stack:
            for i,v  in enumerate(self.stack[:k]):
                self.stack[i] += val
        
            	      
		

运行结果

Runtime: 132 ms, faster than 27.21% of Python online submissions for Design a Stack With Increment Operation.
Memory Usage: 14.5 MB, less than 9.56% of Python online submissions for Design a Stack With Increment Operation.

原题链接：leetcode.com/problems/de…

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