【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; } }转载于:https://www.cnblogs.com/bhlsheji/p/4296177.html
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