Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. push(x) -- Push element x onto stack. pop() -- Removes the element on top of the stack. top() -- Get the top element. getMin() -- Retrieve the minimum element in the stack. Example: MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> Returns -3. minStack.pop(); minStack.top(); --> Returns 0. minStack.getMin(); --> Returns -2. Subscribe to see which companies asked this question.
题意：设计一个有MIN函数的栈。
public class MinStack {
private Stack<Integer> stackData;
private Stack<Integer> stackMin;
/** initialize your data structure here. */
public MinStack() {
stackData=new Stack<Integer>();
stackMin=new Stack<Integer>();
}
public void push(int x) {
if(stackMin.isEmpty()){
stackMin.push(x);
}else if(x<=stackMin.peek()){
stackMin.push(x);
}
stackData.push(x);
}
/////这里貌似有点问题。。。。//包装类比较。。。。
public void pop() {
if(stackData.peek().equals(stackMin.peek())){
stackMin.pop();
}
stackData.pop();
}
public int top() {
return stackData.peek();
}
public int getMin() {
return stackMin.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();
*/PS:定义一个辅助栈。用于存放每一步的最小值。参见左老师的书P1。
