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 <= 10001 <= x <= 10001 <= k <= 10000 <= 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);
*/