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.

A:

Tricky的地方就是： 每次inc 只记录一个位置，然后每次退出一个位置的时候，把其值加到前一个上。

class CustomStack {
public:
    CustomStack(int maxSize) {
        capacity = maxSize;
        IncVal = vector<int>(maxSize,0);
    }
    
    void push(int x) {
        if(values.size() >= capacity)
            return;
        values.push(x);
    }
    
    int pop() {
        if(values.empty())
            return -1;
        int res = values.top()+IncVal[values.size()-1];
        if(values.size() > 1){
            IncVal[values.size()-2] += IncVal[values.size()-1];
        }
        IncVal[values.size()-1] = 0;
        values.pop();
        return res;
    }
    
    void increment(int k, int val) {
        if(values.size()>0 && k > 0){
            IncVal[min(k-1,int(values.size()-1))] += val;
        }
    }
private:
    stack<int> values;
    vector<int> IncVal;
    int capacity =0;
};

/**
 * Your CustomStack object will be instantiated and called as such:
 * CustomStack* obj = new CustomStack(maxSize);
 * obj->push(x);
 * int param_2 = obj->pop();
 * obj->increment(k,val);
 */