设计一个支持 push,pop,top 操作,并能在常数时间内检索到最小元素的栈。
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.
解法1:
class MinStack {
Stack<Integer> stack=null;
Stack<Integer> minStack=null;
/** initialize your data structure here. */
public MinStack() {
stack=new Stack<>();
minStack=new Stack<>();
}
public void push(int x) {
if (stack.isEmpty()&&minStack.isEmpty()){
stack.push(x);
minStack.push(x);
}else {
int y=minStack.peek();
stack.push(x);
if (y<x){
minStack.push(y);
}else {
minStack.push(x);
}
}
}
public void pop() {
stack.pop();
minStack.pop();
}
public int top() {
return stack.get(stack.size()-1);
}
public int getMin() {
return minStack.peek();
}
}
解法2:class MinStack {
Stack<Integer> stack=null;
/** initialize your data structure here. */
public MinStack() {
stack=new Stack<>();
}
public void push(int x) {
if (stack.isEmpty()){
stack.push(x);
stack.push(x);
}else {
int y=stack.peek();
stack.push(x);
if (y<x){
stack.push(y);
}else {
stack.push(x);
}
}
}
public void pop() {
stack.pop();
stack.pop();
}
public int top() {
return stack.get(stack.size()-2);
}
public int getMin() {
return stack.peek();
}
}