1381. Design a Stack With Increment Operation (M)

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.

题意

实现一个拥有指定操作的栈。

思路

最简单的就是直接照着题目意思实现这个栈。

也存在increment()方法为的解法：[Java/C++/Python] Lazy increment, O(1))

代码实现

class CustomStack {
    private List<Integer> stack;
    private int maxSize;
    public CustomStack(int maxSize) {
        stack = new ArrayList<>();
        this.maxSize = maxSize;
    }
    public void push(int x) {
        if (stack.size() == maxSize) {
            return;
        }
        stack.add(x);
    }
    public int pop() {
        if (stack.size() == 0) {
            return -1;
        }
        return stack.remove(stack.size() - 1);
    }
    public void increment(int k, int val) {
        int pos = Math.min(k, stack.size());
        for (int i = 0; i < pos; i++) {
            stack.set(i, stack.get(i) + val);
        }
    }
}

代码实现 - O(1) increment()

class CustomStack {
    private Deque<Integer> stack;
    private int maxSize;
    private int[] append;
    public CustomStack(int maxSize) {
        stack = new ArrayDeque<>();
        this.maxSize = maxSize;
        append = new int[maxSize];
    }
    public void push(int x) {
        if (stack.size() == maxSize) {
            return;
        }
        stack.push(x);
    }
    public int pop() {
        if (stack.size() == 0) {
            return -1;
        }
        int pos = stack.size() - 1;
        if (pos > 0) {
            append[pos - 1] += append[pos];
        }
        int res = stack.pop() + append[pos];
        append[pos] = 0;
        return res;
    }
    public void increment(int k, int val) {
        int pos = Math.min(k, stack.size()) - 1;
        if (pos >= 0) {
            append[pos] += val;
        }
    }
}