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.

分析：https://leetcode.com/problems/design-a-stack-with-increment-operation/discuss/539716/JavaC%2B%2BPython-Lazy-increment-O(1)

 Use an additional array to record the increment value. inc[i] means for all elements stack[0] ~ stack[i],we should plus inc[i] when popped from the stack. Then inc[i-1]+=inc[i], so that we can accumulate the increment inc[i] for the bottom elements and the following pops.

 1 class CustomStack {
 2     int n;
 3     int[] inc;
 4     Stack<Integer> stack;
 5     public CustomStack(int maxSize) {
 6         n = maxSize;
 7         inc = new int[n];
 8         stack = new Stack<>();
 9     }
10 
11     public void push(int x) {
12         if (stack.size() < n) {
13             stack.push(x);
14         }
15     }
16 
17     public int pop() {
18         int i = stack.size() - 1;
19         if (i < 0) return -1;
20         if (i > 0) inc[i - 1] += inc[i];
21         int res = stack.pop() + inc[i];
22         inc[i] = 0;
23         return res;
24     }
25 
26     public void increment(int k, int val) {
27         int i = Math.min(k, stack.size()) - 1;
28         if (i >= 0) {
29             inc[i] += val;
30         }
31     }
32 }
33 
34 /**
35  * Your CustomStack object will be instantiated and called as such:
36  * CustomStack obj = new CustomStack(maxSize);
37  * obj.push(x);
38  * int param_2 = obj.pop();
39  * obj.increment(k,val);
40  */