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分析

难度 易

来源

https://leetcode.com/problems/min-stack/

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.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

解答

 

  1 package LeetCode;
  2 
  3 import java.util.LinkedList;
  4 import java.util.Stack;
  5 
  6 /**
  7  * Your MinStack object will be instantiated and called as such:
  8  * MinStack obj = new MinStack();
  9  * obj.push(x);
 10  * obj.pop();
 11  * int param_3 = obj.top();
 12  * int param_4 = obj.getMin();
 13  */
 14 public class L155_MinStack {
 15    /*  private LinkedList<Integer> stack=null;
 16     int min=Integer.MAX_VALUE;
 17     public L155_MinStack() {
 18         stack=new LinkedList<Integer>();
 19     }
 20 
 21     public void push(int x) {
 22         if(x<=min){
 23             stack.addLast(min);//把之前的min压栈
 24             min=x;
 25         }
 26         stack.addLast(x);
 27     }
 28 
 29     public void pop() {
 30         if(min==stack.removeLast())
 31             min=stack.removeLast();
 32     }
 33 
 34     public int top() {
 35         return stack.getLast();
 36     }
 37 
 38     public int getMin() {
 39         return min;
 40     }*/
 41    //数组实现
 42       /*实时改变min
 43    Runtime: 76 ms, faster than 46.13% of Java online submissions for Min Stack.
 44     */
 45   /* private Integer[] stack=null;
 46     private int maxSize=10000;
 47     private int top=-1;
 48     int min=Integer.MAX_VALUE;
 49     public L155_MinStack() {
 50         stack=new Integer[maxSize];
 51     }
 52 
 53     public void push(int x) {
 54         if(top==maxSize-1){//栈满
 55             maxSize*=2;
 56             Integer[] temp=stack;
 57             stack=new Integer[maxSize];
 58             for(int i=0;i<temp.length;i++){
 59                 stack[i]=temp[i];
 60             }
 61         }
 62         stack[++top]=x;
 63         if(stack[top]<min)
 64             min=stack[top];
 65     }
 66 
 67     public void pop() {
 68         if(stack[top]==min){
 69             min=Integer.MAX_VALUE;
 70             for(int i=0;i<top;i++)
 71             {
 72                 if(stack[i]<min)
 73                     min=stack[i];
 74             }
 75         }
 76         top--;
 77     }
 78 
 79     public int top() {
 80         return stack[top];
 81     }
 82 
 83     public int getMin() {
 84         return min;
 85     }*/
 86    /*注意最小值的保存,否则后边没法求min
 87    使用java的stack
 88     */
 89     private Stack<Integer> stack=null;
 90     int min=Integer.MAX_VALUE;
 91     public L155_MinStack() {
 92         stack=new Stack<Integer>();
 93     }
 94 
 95     public void push(int x) {//int比Integer快好多啊
 96         if(x<=min){
 97             stack.push(min);//把之前的min压栈
 98             min=x;
 99         }
100         stack.push(x);
101     }
102 
103     public void pop() {
104         if(min==stack.pop())//当前栈顶数字等于最小值,栈顶出栈,最小数字为下方数字,栈顶入栈前的最小数字。不管等不等,pop操作都执行过了
105             min=stack.pop();//把最小值出栈,余下正常栈顶
106     }
107 
108     public int top() {
109         return stack.peek();
110     }
111 
112     public int getMin() {
113         return min;
114     }  
115 }

 

posted on 2018-11-09 22:38  flowingfog  阅读(47)  评论(0编辑  收藏