含记录最小元素的堆栈

来源：互联网发布：郑爽整了哪些部位知乎编辑：程序博客网时间：2024/06/05 09:52

设计一种堆栈(stack)，它能 push，pop，并且能在常数时间内O(1)找到当前栈中的最小元素。

使用double space方式，用空间换时间：

public class Stack{int [][]stack;int top;public Stack(){stack = new int[10][2];top =0;}public void push(int a){if(top == 0)stack[top][1] = a;else if(a<stack[top-1][1])stack[top][1] = a;elsestack[top][1] = stack[top-1][1];stack[top][0] = a;top ++;}public int pop(){if(top>0){top--;return stack[top][0];}elsereturn 0;}public int findMin(){if(top>0)return stack[top-1][1];elsereturn 0;}public static void main(String args[]){Stack s = new Stack();s.push(13);s.push(5);s.push(7);s.push(2);s.push(9);s.push(14);System.out.println(s.findMin());s.pop();System.out.println(s.findMin());s.pop();System.out.println(s.findMin());s.pop();System.out.println(s.findMin());s.pop();System.out.println(s.findMin());s.pop();System.out.println(s.findMin());}}

仅增加一个变量记录最小值，每次pop，push时需重新记录最小值。

public class Stack2 {int []stack;int top = 0;int min = 2147483647;public Stack2(){stack = new int[10];}public void push(int a){if(a<min)min=a;stack[top] = a;top++;}public int pop(){if(top>0){top--;if(top == 0)min=0;else if(stack[top] == min){min = stack[top-1];for(int i=0;i<top-1;i++)if(stack[i]<min)min = stack[i];}return stack[top];}return 0;}public int findMin(){return min;}public static void main(String[] args) {Stack2 s = new Stack2();s.push(13);s.push(5);s.push(7);s.push(2);s.push(9);s.push(14);System.out.println(s.findMin());s.pop();//14System.out.println(s.findMin());s.pop();//9System.out.println(s.findMin());s.pop();//2System.out.println(s.findMin());s.pop();//7System.out.println(s.findMin());s.pop();//5System.out.println(s.findMin());}}