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
。
示例:
输入: ["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"] [[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]] 输出: [null,null,null,2,null,null,null,null,null,103,202,201,-1] 解释:
CustomStack customStack = new CustomStack(3); // 栈是空的 [] customStack.push(1); // 栈变为 [1] customStack.push(2); // 栈变为 [1, 2] customStack.pop(); // 返回 2 --> 返回栈顶值 2,栈变为 [1] customStack.push(2); // 栈变为 [1, 2] customStack.push(3); // 栈变为 [1, 2, 3] customStack.push(4); // 栈仍然是 [1, 2, 3],不能添加其他元素使栈大小变为 4 customStack.increment(5, 100); // 栈变为 [101, 102, 103] customStack.increment(2, 100); // 栈变为 [201, 202, 103] customStack.pop(); // 返回 103 --> 返回栈顶值 103,栈变为 [201, 202] customStack.pop(); // 返回 202 --> 返回栈顶值 202,栈变为 [201] customStack.pop(); // 返回 201 --> 返回栈顶值 201,栈变为 [] customStack.pop(); // 返回 -1 --> 栈为空,返回 -1
提示:
1 <= maxSize <= 1000
1 <= x <= 1000
1 <= k <= 1000
0 <= val <= 100
- 每种方法
increment
,push
以及pop
分别最多调用1000
次
题目解释:用vector数组模拟一下
class CustomStack { int mx; vector<int>v; public: CustomStack(int maxSize) { mx=maxSize; v.clear(); } void push(int x) { if(v.size()<mx)v.push_back(x); } int pop() { if(v.size()>0){ int x=v.back(); v.pop_back(); return x; } return -1; } void increment(int k, int val) { int minx=k<v.size()?k:v.size(); for(int i=0;i<minx;i++) v[i]+=val; } }; /** * 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); */