设计一个支持 push,pop,top 操作,并能在常量时间内检索最小元素的栈。
push(x) -- 将元素x推入栈中。
pop() -- 删除栈顶的元素。
top() -- 获取栈顶元素。
getMin() -- 检索栈中的最小元素。
示例:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.
详见:https://leetcode.com/problems/min-stack/description/
Java实现:
方法一:
class MinStack {
private Stack<Integer> stk;
private Stack<Integer> minStk;
/** initialize your data structure here. */
public MinStack() {
stk=new Stack<Integer>();
minStk=new Stack<Integer>();
}
public void push(int x) {
stk.push(x);
if(minStk.isEmpty()){
minStk.push(x);
}else if(minStk.peek()>=x){
minStk.push(x);
}
}
public void pop() {
int top=stk.pop();
if(minStk.peek()==top){
minStk.pop();
}
}
public int top() {
return stk.peek();
}
public int getMin() {
return minStk.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
方法二:
class MinStack {
private Stack<Integer> stk;
private Stack<Integer> minStk;
/** initialize your data structure here. */
public MinStack() {
stk=new Stack<Integer>();
minStk=new Stack<Integer>();
}
public void push(int x) {
stk.push(x);
if(minStk.isEmpty()){
minStk.push(x);
}else if(minStk.peek()>=x){
minStk.push(x);
}else{
minStk.push(minStk.peek());
}
}
public void pop() {
stk.pop();
minStk.pop();
}
public int top() {
return stk.peek();
}
public int getMin() {
return minStk.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/