【leetcode】1381. Design a Stack With Increment Operation

题目如下：

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.

解题思路：本题对性能的要求不高，调用inc的时候把所有符合条件的元素都加上指导的值就可以。

代码如下：

class CustomStack(object):

    def __init__(self, maxSize):
        """
        :type maxSize: int
        """
        self.max_size = maxSize
        self.val_list = []

    def push(self, x):
        """
        :type x: int
        :rtype: None
        """
        if len(self.val_list) < self.max_size:
            self.val_list.append(x)


    def pop(self):
        """
        :rtype: int
        """
        if len(self.val_list) == 0:
            return -1
        return self.val_list.pop(-1)

    def increment(self, k, val):
        """
        :type k: int
        :type val: int
        :rtype: None
        """
        for i in range(min(k,len(self.val_list))):
            self.val_list[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)