LeetCode -- Min Stack

Question:

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.

Analysis:

设计一个栈，它能够支持push, pop, top操作，并且能够在常数时间内检索到最小的元素。

solution1: 利用Java内置的Stack类，并且设置一个格外的栈，当每次有元素进栈时都判断是否是额外栈的最小元素。这样当求最小元素时直接peek额外栈的栈顶元素即可。

solution2: 不适用Java内置的Stack类，每个栈节点维护一个最小值（是当前节点至栈底的最小值），其余操作与上面相同，只不过push，pop等基本操作要自己简单的实现一下。

Answer:

Solution1：

class MinStack {
private Stack<Integer> stack = new Stack<Integer>();
private Stack<Integer> minStack = new Stack<Integer>();
public void push(int x) {
if(minStack.isEmpty() || x <= minStack.peek()) //if x is the minimum value
minStack.push(x);
stack.push(x);
}

public void pop() {
if(stack.peek().equals(minStack.peek())) //if the pop value is the minimum
minStack.pop();
stack.pop();
}

public int top() {
return stack.peek();
}

public int getMin() {
return minStack.peek();
}
}

Solution2：

class MinStack {
StackNode top = null;
public void push(int x) {
if(top == null) {
top = new StackNode(x);
top.min = x;
top.next = null;
}
else {
StackNode node = new StackNode(x);
if(x < top.min)
node.min = x;
else node.min = top.min;
node.next = top;
top = node;
}
}

public void pop() {
if(top != null)
top = top.next;
}

public int top() {
if(top != null)
return top.val;
else return 0;
}

public int getMin() {
if(top != null)
return top.min;
else return 0;
}
}
class StackNode {
int val;
int min; //store the min utile this node
StackNode next;
public StackNode(int x) {
val = x;
}
}