题目描述

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.

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：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。

算法1

题解：我们需要一个数组来辅助记录每一个位置的数字被加上了多少，这样出栈的时候直接是栈中元素和当前辅助数组元素的和。如果直接记录的话，时间复杂度是$O(k)$的。我们可以简化成$O(1)$的操作：每次只在辅助数组第$k$个位置上加上val，等到这个元素出栈的时候，我们将该辅助数组上的arr[k]加到arr[k - 1]上，同时将arr[k]清零。为了描述清晰，我这里没有使用数组模拟栈。

class CustomStack {
public:
    vector<int> inc_arr;
    stack<int> st;
    int m_size,cnt = 0;
    CustomStack(int maxSize) {
        m_size = maxSize;
        inc_arr.resize(maxSize + 1,0);
    }

    void push(int x) {
        if(cnt < m_size){
            cnt ++;
            st.push(x);
        }
    }

    int pop() {
        if(cnt == 0)
            return -1;
        int p = st.top();
        st.pop();
        if(inc_arr[cnt] != 0){
            inc_arr[cnt - 1] += inc_arr[cnt];
            p += inc_arr[cnt];
            inc_arr[cnt] = 0;
        }
        cnt --;
        return p;
    }

    void increment(int k, int val) {
        inc_arr[min(k,cnt)] += val;
    }
};