# Crackingcodinginterview (3.2) stack implements the min function of constant complexity

Source: Internet
Author: User
3.2 Howwouldyoudesignastackwhich, inadditiontopushandpop, alsohasafunctionminwhichreturnstheminimumelement? Push, popandminshouldalloperateinO (1) time.1.Stack1, push (), pop () time complexity: O (n) 2. Stack2, with Stack3, full

3.2 How wocould you design a stack which, in addition to push and pop, also has a function min which returns the minimum element? Push, pop and min shoshould all operate in O (1) time. 1. stack1, push (), pop () time complexity: O (n) 2. stack2, with Stack3, full

3.2 How wocould you design a stack which, in addition to push and pop, also has a function min which returns the minimum element? Push, pop and min shoshould all operate in O (1) time.

1. Stack1, push (), pop () time complexity: O (n)

2. Stack2 and Stack3 meet requirements. Stack3 is optimized to eliminate some redundancy.

3. Stack features. When the current element is popped up, the minimum value of the current element will not change.

```import java.util.Stack;class Stack1{private Node top = null;private Node first = null;class Node{int val;Node next;Node min;public Node(int val){this.val = val;this.next = null;this.min = null;}}//time complexity:O(n) public void push(int val){if(top != null){Node n = new Node(val);n.next = top;top  = n;if(first.val < val){//time complexity:O(n)Node p = null;for(p = first;;p = p.min)if(p.min == null || val < p.min.val){n.min = p.min;p.min = n;break;}}else{n.min = first;first = n;}}else{top = new Node(val);first = top;}}//time complexity:O(n) public int pop(){if(top != null){Node n = top;top = top.next;Node p = null;if(!n.equals(first)){for(p = first;!n.equals(p.min);p = p.min);p.min = p.min.min;}else{first = first.min;}return n.val;}else{return Integer.MIN_VALUE;}}//time complexity:O(1)public int min(){if(first != null)return first.val;elsereturn Integer.MAX_VALUE;}public boolean empty(){if(top == null)return true;elsereturn false;}}class Stack2{private Node top;class Node{int val;int min;Node next;public Node(int val, int min){this.val = val;this.min = min;this.next = null;}}public void push(int val){if(top != null){Node n = new Node(val, val < top.min ? val : top.min);n.next = top;top = n;}else{top = new Node(val, val);}}public int pop(){if(top != null){Node n = top;top = top.next;return n.val;}else{return Integer.MIN_VALUE;}}public int min(){if(top != null)return top.min;elsereturn Integer.MAX_VALUE;}public boolean empty(){if(top == null)return true;elsereturn false;}}class Stack3{private Node top = null;private Stack

s = new Stack

();class Node{int val;Node next;public Node(int val){this.val = val;this.next = null;}}public void push(int val){if(top != null){Node n = new Node(val);n.next = top;top = n;if(s.peek() >= val)s.push(val);}else{top = new Node(val);s.push(val);}}public int pop(){if(top != null){Node n = top;top = top.next;if(n.val == s.peek())s.pop();return n.val;}elsereturn Integer.MIN_VALUE;}public int min(){if(top == null)return Integer.MAX_VALUE;elsereturn s.peek();}public boolean empty(){if(top == null)return true;elsereturn false;}}public class Solution{public static void main(String[] args){int[] A = {23, 11, 12, 34, 10, 12, 7, 45, 21,12, 6, 12, 5, 85, 4, 3, 2, 1};//test for Stack1Stack1 stack1 = new Stack1();for(int i=0;i < A.length;i++){stack1.push(A[i]);}while(!stack1.empty()){System.out.print(stack1.pop() + "[" +stack1.min() + "]" + " ");}System.out.println();//test for Stack2Stack2 stack2 = new Stack2();for(int i=0;i < A.length;i++){stack2.push(A[i]);}while(!stack2.empty()){System.out.print(stack2.pop() + "[" +stack2.min() + "]" + " ");}System.out.println();//test for Stack3Stack3 stack3 = new Stack3();for(int i=0;i < A.length;i++){stack3.push(A[i]);}while(!stack3.empty()){System.out.print(stack3.pop() + "[" +stack3.min() + "]" + " ");}System.out.println();}}

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