155. Min Stack.md

155. Min Stack

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Difficulty : Easy

Related Topics : Stack、Design

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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 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

Constraints:

• Methods pop, top and getMin operations will always be called on non-empty stacks.

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Solution

• mine
  • Java
    • Runtime: 3 ms, faster than 100.00%, Memory Usage: 41.6 MB, less than 12.63% of Java online submissions

      LinkedList<Node> list;
      int min;
      
      public MinStack() {
          list = new LinkedList<>();
          min = Integer.MAX_VALUE;
      }
      
      public void push(int x) {
          min = Math.min(x, min);
          list.add(new Node(x, min));
      }
      
      public void pop() {
          list.removeLast();
          if(list.size() > 0){
              min = list.getLast().m;
          }else{
              min = Integer.MAX_VALUE;
          }
      }
      
      public int top() {
          return list.getLast().val;
      }
      
      public int getMin() {
          return list.getLast().m;
      }
      
      class Node {
          int val;
          int m;
      
          public Node(int val, int m) {
              this.val = val;
              this.m = Math.min(m, val);
          }
      }
      

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