【题目】
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.
【用Java内置的Stack实现】
来自：https://oj.leetcode.com/discuss/15659/simple-java-solution-using-two-build-in-stacks?state=edit-15691&show=15659#q15659
class MinStack {
// stack: store the stack numbers
private Stack<Integer> stack = new Stack<Integer>();
// minStack: store the current min values
private Stack<Integer> minStack = new Stack<Integer>();
public void push(int x) {
// store current min value into minStack
if (minStack.isEmpty() || x <= minStack.peek())
minStack.push(x);
stack.push(x);
}
public void pop() {
// use equals to compare the value of two object, if equal, pop both of them
if (stack.peek().equals(minStack.peek()))
minStack.pop();
stack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}
【分析】
这道题的关键之处就在于 minStack 的设计，push() pop() top() 这些操作Java内置的Stack都有，不必多说。
我最初想着再弄两个数组，分别记录每一个元素的前一个比它大的和后一个比它小的，想复杂了。
第一次看上面的代码，还认为它有问题，为啥仅仅在 x<minStack.peek() 时压栈？假设，push(5), push(1), push(3) 这样minStack里不就仅仅有5和1，这样pop()出1后， getMin() 不就得到5而不是3吗？事实上这样想是错的，由于要想pop()出1之前，3就已经被pop()出了。.
minStack 记录的永远是当前全部元素中最小的，不管 minStack.peek() 在stack 中所处的位置。
【不用内置Stack的实现】
来自：https://oj.leetcode.com/discuss/15651/my-java-solution-without-build-in-stack
class MinStack {
Node top = null;
public void push(int x) {
if (top == null) {
top = new Node(x);
top.min = x;
} else {
Node temp = new Node(x);
temp.next = top;
top = temp;
top.min = Math.min(top.next.min, x);
}
}
public void pop() {
top = top.next;
return;
}
public int top() {
return top == null ? 0 : top.val;
}
public int getMin() {
return top == null ? 0 : top.min;
}
}
class Node {
int val;
int min;
Node next;
public Node(int val) {
this.val = val;
}
}